Algebra - Homepage | Wiley · number and algebra 3.2 Using pronumerals ... a 3x2 b x + 5 + 2 C x2...

3.1 OverviewWhy learn this?Algebra relates to you and the world around you. It is part of everyday life and you will be using it without knowing it. If you want to be a scientist or engineer, you will certainly need algebra. If you want to do well in Maths, you will need to study algebra. When you learn algebra you learn a different way of thinking. It helps with problem solving, decision making, reasoning and creative thinking.

What do you know? 1 THInK Use a thinking tool such as a concept map to list

what you know about algebra.2 PaIr Share what you know with a partner and then with a

small group.3 SHare As a class, create a thinking tool such as a concept map to

show your class’s knowledge of algebra.

Learning sequence3.1 Overview3.2 Using pronumerals3.3 Algebra in worded problems3.4 Simplifying algebraic expressions3.5 Expanding brackets3.6 Expansion patterns3.7 Further expansions3.8 The highest common factor3.9 The highest common binomial factor

3.10 Applications3.11 Review ONLINE ONLY

Algebra

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number and algebra

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WaTCH THIS vIdeOThe story of mathematics:One small step for man . . .

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number and algebra

3.2 Using pronumerals • A pronumeral is a letter or symbol which stands for a number. Algebra is the branch of

mathematics where we use and manipulate pronumerals.

Expressions • In algebra an expression is a group of terms in which each term is separated by a + sign

or a − sign. • Here are some important things to note about the expression 5x − 3y + 2:

– The expression has 3 terms and is called a trinomial. – The pronumerals x and y may take different values and are called variables. – The term 2 has only one value and is called a constant. – The coeffi cient (number in front) of x is 5, and the coeffi cient of y is −3.

• An expression with 2 terms is called a binomial, and an expression with 1 term is called a monomial. A polynomial is an expression with many terms.

For the expression 6x − 3xy + z + 2 + x2z + y2

7, determine:

a the number of terms b the coeffi cient of the second termc the coeffi cient of the last term d the constant term e the term with the smallest coeffi cient f the coeffi cient of x2z.THInK WrITe

a Count the number of terms. a There are 6 terms.

b The second term is −3xy. The number part is the coeffi cient.

b The coeffi cient of the second term is −3.

c The last term is y2

7. This can be rewritten as 1

7× y2. c The coeffi cient of the last

term is 17.

d The constant is the term with no pronumeral. d The constant term is 2.

e Identify the smallest coeffi cient and write the whole term to which it belongs.

e The term with the smallest coeffi cient is −3xy.

f x2z can be written as 1x2z. f The coeffi cient of x2z is 1.

Substitution • We can evaluate (fi nd the value of) an algebraic expression if we replace the pronumerals

with their known values. • This process is called substitution. • Consider the expression 4x + 3y. If we substitute the values x = 2 and y = 5, then the

expression takes the value4 × 2 + 3 × 5 = 8 + 15

= 23 • Rather than showing the multiplication signs, it is common in mathematics to write the

substituted values in brackets. We would write the example above as:

4x + 3y = 4(2) + 3(5) = 8 + 15 = 23

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Topic 3 • Algebra 47

Evaluate the following terms for the values x = 4 and x = −4.a 3x2 b −x2 c 5 − x

THInK WrITe

a 1 Substitute 4 for x. a If x = 4: 3x2 = 3(4)2

2 Only the x term is squared. = 3 × 16

3 Evaluate. = 48

4 Repeat for x = −4. If x = −4: 3x2 = 3(−4)2

= 3 × 16= 48

b 1 Substitute 4 for x. b If x = 4: −x2 = −(4)2

2 Evaluate. = −16

3 Repeat for x = −4. If x = −4: −x2 = −(−4)2

= −16

c 1 Substitute 4 for x. c If x = 4: 5 − x = 5 − 4

2 Evaluate. = 1 = 1

3 Repeat for x = −4. If x = −4: 5 − x = 5 − (−4) = 5 + 4 = 9 = 3

If x = 3 and y = −2, evaluate the following expressions.a 3x + 2y b 5xy − 3x + 1 c 2x2 + y2

THInK WrITe

a 1 Write the expression. a 3x + 2y

2 Substitute x = 3 and y = −2. = 3(3) + 2(−2)

3 Evaluate. = 9 − 4 = 5

b 1 Write the expression. b 5xy − 3x + 12 Substitute x = 3 and y = −2. = 5(3)(−2) − 3(3) + 13 Evaluate. = −30 − 9 + 1

= −38

c 1 Write the expression. c 2x2 + y2

2 Substitute x = 3 and y = −2. = 2(3)2 + (−2)2

3 Evaluate. = 18 + 4 = 22

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48 Maths Quest 9

Substitution into formulas • A formula is a mathematical rule, usually written using pronumerals. • We know that the formula for the area of a rectangle is given by:

A = lwwhere A represents the area of the rectangle, l represents the length and w represents the width.

• If a rectangular kitchen tile has length l = 20 cm and width w = 15 cm, we can substitute these values into the formula to fi nd its area.

A = lw= 20 × 15= 300 cm2

• If we are given the area of the tile to be 400 cm2 and the width to be 55 cm, then we can calculate its length.

A = lw400 = l × 55

l = 40055

= 8011

≈ 7.3 cm (to 1 d.p.)

The formula for the voltage in an electrical circuit can be found using the formula known as Ohm’s Law:

V = IRwhere I = current in amperes

R = resistance in ohmsV = voltage in volts.

a Calculate V when: i I = 2 amperes, R = 10 ohmsii I = 20 amperes, R = 10 ohms.

b Calculate I when V = 300 volts and R = 600 ohms.

THInK WrITe

a i 1 Write the formula. a i V = IR2 Substitute I = 2 and R = 10. = (2)(10)

3 Evaluate and express the answer in the correct units.

= 20The voltage is 20 volts.

ii 1 Write the formula. ii V = IR2 Substitute I = 20 and R = 10. = (20)(10)


= 200The voltage is 200 volts.



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b 1 Write the formula. b V = IR2 Substitute V = 300 and R = 600. 300 = I(600)


I = 300600

= 0.5The current is 0.5 amperes.

Note: Even if we know nothing about volts, amperes and ohms, we can still do the calculation.

Exercise 3.2 Using pronumerals IndIvIdual PaTHWaYS

⬛ PraCTISeQuestions:1, 2a–h, 3a–f, 4–6, 8, 9a–g, 10–12, 20

⬛ COnSOlIdaTeQuestions:1, 2e–l, 3c–h, 4–8, 9d–k, 13–16, 18, 20

⬛ maSTerQuestions:1, 2i–o, 3e–j, 4–6, 8, 9j–o, 15–20

FluenCY

1 Find the coeffi cient of each of the following terms.a 3x b 7a c −2m d −8q e w

f −n g x3

h y2

i −at4

j −r2

92 State the coeffi cient of the x term in each of the following expressions.

a 6x − 3y b 5 + 7x c 5x2 + 3x − 2d −7x2 − 2x + 4 e 3x − 2x2 f −9x2 − 2x

g 5x2 + 3 − 7x h −11x + 5 − 2x2 i 2ax3 − x2

+ x2

j 1 − 2x3

− bx2 k x2

6 + 5

7 + x

4 l x3 + x + 4

m −3x − 4bx + 6 n 4cx2 − 2x + 4ax o 2x2 − 53 WE1 For each expression below, write:

i the number of terms ii the coeffi cient of the fi rst termiii the constant term iv the term with the smallest coeffi cient.

a 5x2 + 7x + 8 b −9m2 + 8m − 6c 5x2y − 7x2 + 8xy + 5 d 9ab2 − 8a − 9b2 + 4e 11p2q2 − 4 + 5p − 7q − p2 f −9p + 5 − 7q2 + 5p2q + qg 4a − 2 + 9a2b2 − 3ac h 5s + s2t + 9 + 12t − 3ui −m + 8 + 5n2m + m2 + 2n j 7c2d + 5d2 + 14 − 3cd2 − 2e

4 WE2 Evaluate the following expressions for x = 1, x = 0 and x = −2.a 5x + 2 b x2 + 5x + 2 c x (x + 2)

d x2 + x −2 e x2 f 2x

5 MC a If x = −3, then the value of −5x − 3 is:a −18 b 12 C 30 d 18

b An expression that is a trinomial is:a 3x2 b x + 5 + 2 C x2 − 7 d x2 + x + 2

reFleCTIOnWhich letters (pronumerals) should you avoid using when writing algebraic expressions?

⬛ ⬛ ⬛ Individual pathway interactivity int-4480

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50 Maths Quest 9

c The expression (x − 6) cannot be evaluated if x is equal to:a 2 b 6 C 7 d 10

d In the expression 5x2 − 3xy + 0.5x − 0.3y − 5, the smallest coefficient is:a 5 b −3 C 0.5 d −0.3

6 WE3 Find the value of the following expressions if x = 2, y = −1 and z = 3.a 2x b 3xy c 2y2z

d 14x

e 6(2x + 3y − z) f x2 − y2 + xyz

7 If x = 4 and y = −3, evaluate the following expressions.a 4x + 3y b 3xy − 2x + 4 c x2 − y2

underSTandIng

8 WE4 The change in the voltage in an electrical circuit can be found using the formula known as Ohm’s Law: V = IR, where I = current in amperes, R = resistance in ohms and V = voltage in volts.a Calculate V when:

i I = 4, R = 8 ii I = 25, R = 10.b Calculate R when:

i V = 100, I = 25 ii V = 90, I = 30. 9 Evaluate each of the following by substituting the given values into each formula.

a If A = bh, find A when b = 5 and h = 3.

b If d = mv

, find d when m = 30 and v = 3.

c If A = 12xy, find A when x = 18 and y = 2.

d If A = 12 (a + b)h, find A when h = 10, a = 7 and b = 2.

e If V = AH3

, find V when A = 9 and H = 10.

f If v = u + at, find v when u = 4, a = 3.2 and t = 2.1.g If t = a + (n − 1)d, find t when a = 3, n = 10 and d = 2.

h If A = 12 (x + y)h, find A when x = 5, y = 9 and h = 3.2.

i If A = 2b2, find A when b = 5.j If y = 5x2 − 9, find y when x = 6.k If y = x2 − 2x + 4, find y when x = 2.l If a = −3b2 + 5b − 2, find a when b = 4.m If s = ut + 1

2at2, find s when u = 0.8, t = 5 and a = 2.3.

n If F = mp

r2, find F correct to 2 decimal places, when m = 6.9, p = 8

and r = 1.2.o If C = πd, find C correct to 2 decimal places if d = 11.

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10 The area of a triangle is given by the formula A = 12bh, where b is the length of the base

and h is the perpendicular height of the triangle.a Show that the area is 12 cm2 when b = 6 cm and h = 4 cm.b What is h if A = 24 cm2 and b = 4 cm?


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11 The formula to convert degrees Fahrenheit (F) to degrees

Celsius (C) is C = 59(F − 32).

a Find C when F = 59.b Show that when Celsius (C) is 15, Fahrenheit (F) is 59.

12 The length of the hypotenuse of a right-angled triangle (c) can

be found using the formula, c = a2 + b2, where a and b are

the lengths of the other two sides.a Find c when a = 3 and b = 4.b Show that when a = 5 and c = 13, b = 12.

13 If the volume of a prism (V) is given by the formula V = AH, where A is the area of the cross-section and H is the height of the prism, determine:a V when A = 7 cm2 and H = 9 cmb H when V = 120 cm3 and A = 30 cm2.

14 Using E = F + V − 2 where F is the number of faces on a prism, E is the number of edges and V is the number of vertices, calculate:a E if F = 5 and V = 7b F if E = 10 and V = 2.

15 The kinetic energy (E) of an object is found by using the formula E = 1

2mv2

where m is the mass and v is the velocity of the object.a Determine E when m = 3 and v = 3.6.b Show that when E = 25 and v = 5,

m = 2.

16 The volume of a cylinder (v) is given by v = πr2h, where r is the radius in centimetres and h is the height of the cylinder in centimetres.a Determine v correct to 2 decimal places if r = 7 and h = 3.b Determine h correct to 2 decimal places if v = 120 and r = 2.

17 The surface area of a cylinder (S) is given by S = 2πr(r + h) where r is the radius of the circular end and h is the height of the cylinder.a Calculate S (to 2 decimal places) if r = 14 and h = 5.b Show that for a cylinder of surface area 240 units2 and radius 5 units, the height is

2.64 units, correct to 2 decimal places.

PrOblem SOlvIng

18 a In a magic square, every row, column and diagonal adds to the same number. This number is called the magic number for that square. Complete the unfinished magic square by first finding an algebraic expression for the magic number.

b Make three different magic squares by substituting different values for g and d in the completed magic square from part a.

c Calculate the magic number for each magic square.

2d + 2g

d + g

d + 2g0


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52 Maths Quest 9

19 A fl at, rectangular board is built by gluing together a number of square pieces of the same size. The board is m squares wide and n squares long. In terms of m and n, write expressions for:a the total number of squaresb the number of completely surrounded squaresc the perimeter of the fi gure, given that each square is 1 unit by 1 unit.

20 Which one of the following eight expressions is not equivalent to the others? x − y + z z − y + x z − (y − x) −y + x + z z + x − y x − (y − z) y − (z − x) x + z − y

3.3 Algebra in worded problems • An important skill in algebra is to be able to convert worded expressions and sentences

into algebraic expressions. • The fi rst step in converting a worded statement into an algebraic expression is to assign

pronumerals to the unknown quantities.

Write an algebraic expression for each of the following, choosing an appropriate pronumeral if necessary.a The number that is 6 more than Ben’s ageb The total value of some $10 notesc The total cost of 8 adults’ and 3 children’s train ticketsd The product of a and we w less than aTHInK WrITe

a 1 Ben’s age is the unknown quantity.Note: You must not say ‘Let Ben = b’; b is a number.

a Let Ben’s age = b years.

2 Write an expression for Ben’s age plus 6. b + 6b 1 The number of notes is the unknown quantity. b Let n = the number of $10 notes.

2 The value is 10 times the number of notes. Total value = 10n dollars

c 1 The cost of tickets is the unknown quantity. c Let an adult’s ticket cost a dollars and a child’s ticket cost c dollars.Total cost = 8a + 3c dollars.

2 The adults’ tickets cost 8 × a and the children’s tickets cost 3 × c.

d ‘Product’ means to multiply. d a × w e w is subtracted from a. e a − w



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Exercise 3.3 Algebra in worded problems IndIvIdual PaTHWaYS

⬛ PraCTISeQuestions:1–5, 7, 12, 14

⬛ COnSOlIdaTeQuestions:3–8, 10, 12, 14, 15

⬛ maSTerQuestions:4–11, 13, 15, 16

FluenCY

1 Jacqueline studies 5 more subjects than Helena. How many subjects does Jacqueline study if:a Helena studies 6 subjects? b Helena studies x subjects?c Helena studies y subjects?

2 Dianne and Angela walk home from school together. Dianne’s home is 2 km further from school than Angela’s home. How far does Dianne walk if Angela’s home is:a 1.5 km from school b x km from school?

3 Lisa watched television for 2.5 hours today. How many hours will she watch tomorrow if she watches:a 1.5 hours more than she watched todayb t hours more than she watched todayc y hours fewer than she watched today?

underSTandIng

4 WE5 Write an algebraic expression for each of the following questions.a If it takes 10 minutes to iron a single shirt, how long would it take to iron all of

Anthony’s shirts?b Ross has 30 dollars more than Nick. If Nick has N dollars, how much money does

Ross have?c In a game of Aussie rules, Luciano kicked 4 more goals than he kicked behinds.

i How many behinds did Luciano score if g is the number of goals kicked?ii How many points did Luciano score?

(Remember: 1 goal scores 6 points, 1 behind scores 1 point.)

reFleCTIOnWhy is it important to de� ne pronumerals or variables in terms of what they represent?


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5 Jeff and Chris play Aussie Rules football for opposing teams, and Jeff’s team won when the two teams played one another.a How many points did Jeff’s team

score if they kicked: i 14 goals and 10 behindsii x goals and y behinds?

b How many points did Chris’s team score if his team kicked: i 10 goals and 6 behindsii p goals and q behinds?

c How many points did Jeff’s team win by if: i Chris’s team scored 10 goals and 6 behinds, and Jeff’s team scored 14 goals and

10 behinds?ii Chris’s team scored p goals and q behinds, and Jeff’s team scored x goals and

y behinds?6 Yvonne’s mother gives her x dollars for each school subject she passes. If she passes

y subjects, how much money does she receive?7 Brian buys a bag containing x Smarties.

a If he divides them equally among n people, how many does each person receive?

b If he keeps half the Smarties for himself and divides the remaining Smarties equally among n people, how many does each person receive?

8 A piece of licorice is 30 cm long.a If David cuts d cm off, how much licorice remains?b If David cuts off 1

4 of the remaining licorice, how much licorice has been cut off?

c How much licorice remains now?

reaSOnIng

9 One-quarter of a class of x students plays tennis on the weekend. One-sixth of the class plays tennis and also swims on the weekend.a Write an expression to

represent the number of students playing tennis on the weekend.

b Write an expression to represent the number of students playing tennis and also swimming on the weekend.

c Show that the number of students playing only tennis on the weekend is x

12.


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10 During a 24-hour period, Vanessa uses her computer for c hours. Her brother Darren uses it for 1

7 of the remaining time.

a For how long does Darren use the computer?b Show that the total number of hours that Vanessa and Darren use the computer

during a 24-hour period can be expressed as 6c + 24

7.

11 Marty had a birthday party last weekend and invited n friends. The table below indicates the number of friends at Marty’s party at certain times during the evening. Everybody arrived by 8.30 pm and had all left the party by 11 pm.a Show that 1 person arrived between 7.00 pm and 7.30 pm.b Between which times were the most friends present at the party?c How many friends were invited but did not arrive?d Prove that 24 friends were invited in total.e Between which times did the most friends arrive?

Time Number of friends

7.00 pm n − 24

7.30 pm n − 23

8.00 pm n − 88.30 pm n − 59.00 pm n − 59.30 pm n − 7

10.00 pm n − 12

10.30 pm n − 18

11.00 pm n − 24

PrOblem SOlvIng

12 A child builds a pyramid as shown.a Find a rule which gives the number of blocks on the

bottom layer, b, for a tower that is h blocks high.b If the child wishes the tower to be 10 blocks high, how

many blocks should she begin with on the bottom layer? 13 If a circle has three equally spaced dots (A, B and C) on its

circumference, only 1 triangle can be formed by joining the dots. Investigate to determine the number of triangles that could be formed by 4, 5, … equally spaced dots on the circumference.

From your investigation, can you see a pattern in the number of triangles (t) formed for a circle with d equally spaced dots on its circumference? Is it possible to express this as a formula?

C

BA BA

CD


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56 Maths Quest 9

14 A shop owner is ordering vases from a stock catalogue. Red vases cost $4.20 and clear vases cost $5.70. If 25 vases cost $120, how many of each type of vase did she buy?

15 a Investigate what the variables in the formula E = mc2 mean. b Prepare a presentation about Albert Einstein and present

this to the class.

16 a Transpose the equation T = x − 3x −1

to make x the subject.

b What is the value of T when x = 1? Explain. c What is the value of x when T = 1? Explain.

3.4 Simplifying algebraic expressions • When we simplify an expression, we write it with the smallest number of terms, without

using × or ÷ signs.

Addition and subtraction of like terms • Like terms have identical pronumeral parts.

For example: 3x and x are like terms, but 3x2 and x are not. 5a2b and −2ba2 are like terms, but 5a2b and −2ab2 are not. 6x + 5x can be simplifi ed (to 11x), but 6x + 5 cannot. • When there is more than one pronumeral in a term, it is preferable to write them in

alphabetical order. For example, 3xy is preferable to 3yx, although these are identical terms. This makes it easy to recognise like terms.

Simplify the following expressions.a x + 2x + 3x − 5 b 9a2b + 5ba2 + ab2 c −12 − 4c2 + 10 + 5c2

THInK WrITe

a 1 Write the expression. a x + 2x + 3x − 52 Collect the like terms: x, 2x, 3x. = 6x − 5

b 1 Write the expression. b 9a2b + 5ba2 + ab2

2 Collect the like terms: 9a2b, 5ba2.

= 14a2b + ab2

c 1 Write the expression. c −12 − 4c2 + 10 + 5c2

2 Collect the like terms. = −12 + 10 − 4c2 + 5c2

3 Simplify. = −2 + c2 or c2 − 2

CHallenge 3.1CHallenge 3.1CHallenge 3.1CHallenge 3.1CHallenge 3.1CHallenge 3.1



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Multiplication and divisionMultiplication • Any two algebraic terms can be multiplied together. • Terms can be multiplied together in any order, so it is easier to deal with the coeffi cients

fi rst and then each pronumeral in turn.3xy × 5y = 3 × x × y × 5 × y

= 3 × 5 × x × y × y= 15 × x × y2

= 15xy2

Simplify the following terms.a 4a × 2b × a b 7ax × −6bx × −2abx

THInK WrITe

a 1 Write the term. a 4a × 2b × a2 Rearrange, writing the coeffi cients fi rst. = 4 × 2 × a × a × b3 Multiply the coeffi cients and

pronumerals separately. = 8a2b

b 1 Write the term. b 7ax × −6bx × −2abx

2 Rearrange, writing the coeffi cients fi rst. = 7 × −6 × −2 × a × a × b × b × x × x × x

3 Multiply the coeffi cients and pronumeral separately.

= 84a2b2x3

Division • Division is performed by writing the terms in fraction form and then cancelling

to simplify.

For example, 10x ÷ 5x = 10x5x

=210x

15x= 2

Simplify the following terms.

a 12xy4xz

b −8ab2 ÷ −16a2b

THInK WrITe

a 1 Write the term. a 12xy4xz

2 Cancel the common factors. = 312 × x × y

14 × x × z

3 Write the answer. = 3yz

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58 Maths Quest 9

b 1 Write the term as a fraction. b −8ab2

−16a2b

2 Cancel the common factors. =1−8 × 1a × b2

b

2 −16 × a

a2 × b1

3 Simplify and write the answer. = 12

× 1a

× b1

= b2a

Exercise 3.4 Simplifying algebraic expressionsIndIvIdual PaTHWaYS

⬛ PraCTISeQuestions:1, 2a–l, 3, 4a–i, 5a–l, 6, 7, 10, 12

⬛ COnSOlIdaTeQuestions:1, 2g–r, 3, 4g–o, 5d–o, 6–13

⬛ maSTerQuestions:1, 2p–z, 3, 4k–r, 5h–r, 6, 8–11, 13–15

FluenCY

1 For each of the following terms, select those terms listed in brackets that are like terms.a 6ab (7a, 8b, 9ab, −ab, 4a2b2)b −x (3xy, −xy, 4x, 4y, −yx)c 3az (3ay, −3za, −az, 3z2a, 3a2z)d x2 (2x, 2x2, 2x3, −2x, −x2)e −2x2y (xy, −2xy, −2xy2, −2x2y, −2x2y2)

f 3x2y5 (3xy, 3x5y2, 3x4y3, −x2y5, −3x2y5)

g 5x2p3w5 (−5x3w5p3, p3x2w5, 5xp3w5, −5x2p3w5, w5p2x3)

h −x2y5z4 (−xy5, −y2z5x4, −x + y + z, 4y5z4x2, −2x2z4y5)

2 WE6 Simplify the following expressions.a 5x + 2x b 3y + 8y

c 7m + 12m d 13q − 2q

e 17r − 9r f −x + 4x

g 5a + 2a + a h 9y + 2y − 3y

i 7x − 2x + 8x j 14p − 3p + 5p

k 2q2 + 7q2 l 5x2 − 2x2

m 6x2 + 2x2 − 3y n 3m2 + 2n − m2

o 9x2 + x − 2x2 p 9h2 − 2h + 3h + 9q −2g2 − 4g + 5g − 12 r −5m2 + 5m − 4m + 15

s 12a2 + 3b + 4b2 − 2b t 6m + 2n2 − 3m + 5n2

u 3xy + 2y2 + 9yx v 3ab + 3a2b + 2a2b − ab

w 9x2y − 3xy + 7yx2 x 4m2n + 3n − 3m2n + 8n

y −3x2 − 4yx2 − 4x2 + 6x2y z 4 − 2a2b − ba2 + 5b − 9a2

reFleCTIOnIs the expression ab the same as ba? Explain.


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doc-6126doc-6126doc-6126doc-6126doc-6126doc-6126



doc-10817doc-10817doc-10817

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SAMPLE E

VALUATIO

N ONLY

number and algebra


3 MC Simplify the following expressions.a 18p − 19p

a p b −p C p2 d −1b 5x2 − 8x + 6x − 9

a 3x − 9 b 3x2 − 9 C 5x2 + 2x − 9 d 5x2 − 2x − 9c 12a − a + 15b − 14b

a 11a + b b 12 C 11a − b d 13a + bd −7m2n + 5m2 + 3 − m2 + 2m2n

a −9m2n + 4m2 + 3 b −9m2n + 8

C −5m2n − 4m2 + 3 d −5m2n + 4m2 + 34 WE7 Simplify the following.

a 3m × 2n b 4x × 5y c 2p × 4qd 5x × −2y e 3y × −4x f −3m × −5ng 5a × 2a h 4y × 5y i 5p × pj m × 7m k 3mn × 2p l −6ab × bm −5m × −2mn n −6a × 3ab o −3xy × −5xy × 2xp 4pq × −p × 3q2 q 4c × −7cd × 2c r −3a2 × −5ab3 × 2ab4

5 WE8 Simplify the following.

a 6x2

b 9m3

c 12y6

d 8m2

e 12m ÷ 3 f 14x ÷ 7 g −21x ÷ 3 h −32m ÷ 8

i 4m8

j 6x18

k 8mn18n

l 16xy12y

m 2

10m n

6ab

12a2b o

28xyz14x

p 70ab2

4b

q 2x2yz8xz

r −7xy2z2 ÷ 11xyz

6 Simplify the following.a 5x × 4y × 2xy b 7xy × 4ax × 2y c x × 4xy × 3yx

d 6x2y

12y2 e

−15x2ab

12b2x2 f

2p3q2

p3q2

g −4a × −5ab2 × 2a h −a × 4ab × 2ba × b i 2a × 2a × 2a × 2a

underSTandIng

7 Jim buys m pens at p cents each and n books at q dollars each.a How much does he spend in:

i dollars ii cents?b What is his change from $20?

8 At a local discount clothing store 4 shirts and 3 pairs of trousers cost $138. If a pair of trousers cost 2.5 times as much as a shirt, determine the cost of each.

reaSOnIng

9 Class 9A were given an algebra test. One of the questions is shown below.

Simplify the following expression: 3ab2

× 4ac6b

× 7c.

Sean, who is a student in class 9A, wrote his answer as 12aabc12b

× 7c. Explain why

Sean’s answer is incorrect, and write the correct answer.


SAMPLE E

VALUATIO

N ONLY

number and algebra

60 Maths Quest 9

PrOblem SOlvIng

10 a Fill in the empty bricks in the following pyramids. The expression in each brick is obtained by adding the expressions in the two bricks below it. i

2a + 3b

3a + b2a – b

ii 8p + 3q

5p – 2q

2p + 3q

b Fill in the empty bricks in the following pyramids. The expression in each brick is obtained by multiplying the expressions in the two bricks below it. i

m2 2m3n

ii 48d2e2

8de

3d

11 For the triangle shown:a write an expression for the perimeter of the triangleb show that the perimeter can be simplifi ed to 43

21 x

c if x is 2 metres, calculate the cost (to the nearest dollar) to frame the triangle using timber which costs $4.50 per metre.

12 For the rectangle shown: a write an expansion for the perimeter of the rectangleb show that the perimeter can be simplifi ed to 64

7w

c if w = 7 metres, calculate the cost (to the nearest cent) to create a wire frame for the rectangle if the wire frame costs $1.57 per metre.

13 A rectangular chocolate block has dimensions x and (4x − 7) as shown. a Write an expression for the perimeter (P) in

the form P = __x + __(4x − 7).b Expand and simplify this equation.c Calculate the perimeter of the chocolate

block when x = 3.d Explain why x cannot equal 1.

x

5x—7

x–3

w

16w—7

4x – 7

x


SAMPLE E

VALUATIO

N ONLY

number and algebra


14 A dog house in the shape of a rectangular prism is to be reinforced with steel edging along its outer edges. The cost of the steel edging is $1.50 per metre. All measurements are in centimetres.

x + 20

x + 40 x + 20

a Write an expression for the total length of the straight edges of the frame.b Expand the expression.c Calculate the cost of steel needed when x = 80 cm.

15 A tile manufacturer produces tiles that have the following side lengths. All measurements are in centimetres.

2x

5x + 123x 4x

a Write an expression for the perimeter of each shape.b Calculate the value of x for which the perimeter of the triangular tile is the same as

the perimeter of the square tile.

3.5 Expanding brackets • What is the area of the large rectangle drawn below? There are two ways of fi nding out.

5 cm 3 cm

4 cm

1. The rectangle can be treated as one large shape. The length of the rectangle is 4 cm, and its width is 5 + 3 cm.

A = l × w= 4(5 + 3)= 4 × 8= 32 cm2

2. The areas of the two small rectangles can be added together.A = 4 × 5 + 4 × 3

= 20 + 12= 32 cm2

• The length of this rectangle is a, and its width is b + c.

b c

a ab ac

b + c

doc-6138doc-6138doc-6138


SAMPLE E

VALUATIO

N ONLY

number and algebra

62 Maths Quest 9

• The area can be found in two different ways.1. A = l × w

= a(b + c)This expression is factorised because it is one number (or factor) multiplied by another. The two factors are a and b + c.

2. The areas of the two small rectangles can be added together.A = ab + ac

This expression is described as expanded or written without brackets. • a(b + c) = ab + ac is called the Distributive Law.

Use two different methods to fi nd the value of the following.a 7(5 + 15) b 10(9 − 1)THInK WrITe

a 1 Method 1: Work out the brackets fi rst, then evaluate.

a 7(5 + 15) = 7 × 20 = 140

2 Method 2: Expand the brackets fi rst, then evaluate.

7(5 + 15) = 7 × 5 + 7 × 15 = 35 + 105 = 140

b 1 Method 1: Work out the brackets fi rst, then evaluate.

b 10(9 − 1) = 10 × 8 = 80

2 Method 2: Expand the brackets fi rst, then evaluate.

10(9 − 1) = 10 × 9 + 10 × −1 = 90 − 10 = 80

• A good way to expand brackets is to draw arrows as you work out each multiplication.

x(x − y) = x2 . . .

x(x − y) = x2 − xy

Expand the following.a 5(x + 3) b −4y(2x − w)THInK WrITe

a 1 Draw arrows to assist in expanding. a 5(x + 3)2 Simplify. = 5 × x + 5 × 3

= 5x + 15

b 1 Draw arrows to assist in expanding. b −4y(2x − w)2 Simplify. = −4y × 2x − 4y × −w

= −8xy + 4wy




SAMPLE E

VALUATIO

N ONLY

number and algebra


Expanding and collecting like terms • With more complicated expansions, like terms may need to be collected after the

expansion of the brackets. First expand the brackets, then collect the like terms.

Expand and simplify by collecting like terms.a 4(x − 4) + 5 b x(y − 2) + 5x c −x(y − z) + 5x d 7x − 6(y − 2x)

THInK WrITe

a 1 Expand the brackets. a 4(x − 4) + 52 Simplify. = 4x − 16 + 5

= 4x − 11

b 1 Expand the brackets. b x(y − 2) + 5x

2 Simplify. = xy − 2x + 5x = xy + 3x

c 1 Expand the brackets. c −x(y − z) + 5x

2 Simplify. (There are no like terms.) = −xy + xz + 5x

d 1 Expand the brackets. d 7x − 6(y − 2x)

2 Simplify. = 7x − 6y + 12x = 19x − 6y

Expanding two brackets • When simplifying an expression containing two (or more) brackets, the steps are

the same. Expand each bracket, and then simplify by collecting like terms.

Expand and simplify the following expressions.a 5(x + 2y) + 6(x − 3y)b 5x(y − 2) − y(x + 3)

THInK WrITe

a 1 Expand each bracket. a 5(x + 2y) + 6(x − 3y)

2 Simplify. = 5x + 10y + 6x − 18y = 11x − 8y

b 1 Expand each bracket. b 5x(y − 2) − y(x + 3)

2 Simplify. = 5xy − 10x − xy − 3y = 4xy − 10x − 3y




SAMPLE E

VALUATIO

N ONLY

number and algebra

64 Maths Quest 9

Expanding binomial factors • Remember, a binomial is an expression containing two terms, such as x + 3 or 2y − z2.

In this section two binomials are multiplied together. • The rectangle below has length a + b and width c + d.

c + d

d

ad

c

aca

bdbcb

a + b

• There are two ways of fi nding the area of the large rectangle.1. A = L × W

= (a + b) × (c + d )= (a + b)(c + d )

This is a factorised expression, where the two factors are (a + b) and (c + d ).2. The areas of the four small rectangles can be added together. A = ac + ad + bc + bd So (a + b)(c + d ) = ac + ad + bc + bd

• There are several methods that can be helpful in remembering how to expand binomial factors. One commonly used method is the FOIL method.

• This method is given the name FOIL because the letters stand for: First — multiply the fi rst term in each bracket. Outer — multiply the 2 outer terms of each bracket. Inner — multiply the 2 inner terms of each bracket. Last — multiply the last term of each bracket.

(a + b)(c + d)II

F

‘FOIL’– First Outer Inner Last

OO

LL

(a + b)(c + d) = ac + ad + bc + bd

OR

(a + b)(c + d)nosenose

eyebrow eyebroweyebrow

mouthmouth

‘eyebrow, eyebrow, nose and mouth’

(a + b)(c + d) = ac + bd + bc + ad

Expand and simplify each of the following expressions.a (x − 5)(x + 3) b (x + 2)(x + 3) c (2x + 2)(2x + 3)

THInK WrITe

a 1 Expand using the FOIL method. a (x − 5)(x + 3)

2 Simplify the expression by collecting like terms. = x2 + 3x − 5x − 15 = x2 − 2x − 15

b 1 Expand the brackets using FOIL. b (x + 2)(x + 3)

2 Simplify the expression by collecting like terms. = x2 + 3x + 2x + 6 = x2 + 5x + 6

c 1 Expand the brackets using FOIL. c (2x + 2)(2x + 3)

2 Simplify the expression by collecting like terms. = 4x2 + 6x + 4x + 6 = 4x2 + 10x + 6



SAMPLE E

VALUATIO

N ONLY

number and algebra


Exercise 3.5 Expanding brackets IndIvIdual PaTHWaYS

⬛ PraCTISeQuestions:1, 2a–d, 3a–d, 4a, c, m, o, 5a, c, g, k, m, 6a, e, i, m, o, 7, 8a, d, g, j, m, 9, 10a, d, g, j, m, 11, 12, 18, 19

⬛ COnSOlIdaTeQuestions:1, 2c–f, 3c–f, 4b, f–j, 5i–p, 6b, d, j, l, p, 7, 8b, e, h, k, n, 9d–f, 10b, e, h, k, n, 13, 14, 18, 19, 22

⬛ maSTerQuestions:1, 2e–h, 3e–h, 4i–p, 5i–p, 6i–p, 7c, 8c, f, i, l, o, 9g–i, 10c, f, i, l, o, 13–17, 20–22

FluenCY

1 WE9 Use two different methods to fi nd the value of these expressions.a 8(10 − 2) b 11(99 + 1) c −5(3 + 1) d 7(100 − 1)

2 WE10 Expand the following expressions. For the fi rst two examples, draw a diagram to represent the expression.a 3(x + 2) b 4(x + 3) c 4(x + 1) d 7(x − 1)e −3(p − 2) f −(x − 1) g 3(2b − 4) h 8(3m − 2)

3 WE10 Expand each of the following. For the fi rst two examples, draw a diagram to represent the expression.a x(x + 2) b a(a + 5) c x(4 + x) d m(7 − m)e 2x(y + 2) f −3y(x + 4) g −b(3 − a) h −6a(5 − 3a)

4 WE11 Expand and simplify by collecting like terms.a 2(p − 3) + 4 b 5(x − 5) + 8c −7(p + 2) − 3 d −4(3p − 1) − 1e 6x(x − 3) − 2x f 2m(m + 5) − 3m g 3x(p + 2) − 5 h 4y(y − 1) + 7i −4p(p − 2) + 5p j 5(x − 2y) − 3y − x k 2m(m − 5) + 2m − 4 l −3p(p − 2q) + 4pq − 1m −7a(5 − 2b) + 5a − 4ab n 4c(2d − 3c) − cd − 5c o 6p + 3 − 4(2p + 5) p 5 − 9m + 2(3m − 1)

5 WE12 Expand and simplify the following expressions.a 2(x + 2y) + 3(2x − y) b 4(2p + 3q) + 2(p − 2q)c 7(2a + 3b) + 4(a + 2b) d 5(3c + 4d ) + 2(2c + d )e −4(m + 2n) + 3(2m − n) f −3(2x + y) + 4(3x − 2y)g −2(3x + 2y) + 3(5x + 3y) h −5(4p + 2q) + 2(3p + q)i 6(a − 2b) − 5(2a − 3b) j 5(2x − y) − 2(3x − 2y)k 4(2p − 4q) − 3(p − 2q) l 2(c − 3d ) − 5(2c − 3d )m 7(2x − 3y) − (x − 2y) n −5(p − 2q) − (2p − q)o −3(a − 2b) − (2a + 3b) p 4(3c + d ) − (4c + 3d )

6 WE12 Expand and simplify the following expressions.a a(b + 2) + b(a − 3) b x(y + 4) + y(x − 2)c c(d − 2) + c(d + 5) d p(q − 5) + p(q + 3)e 3c(d − 2) + c(2d − 5) f 7a(b − 3) − b(2a + 3)g 2m(n + 3) − m(2n + 1) h 4c(d − 5) + 2c(d − 8)i 3m(2m + 4) − 2(3m + 5) j 5c(2d − 1) − (3c + cd )


doc-10819doc-10819doc-10819

doc-10820doc-10820doc-10820

reFleCTIOnExplain why, when expanded, (x + y)(2x + y) gives the same result as (2x + y)(x + y).


SAMPLE E

VALUATIO

N ONLY

number and algebra

66 Maths Quest 9

k −3a(5a + b) + 2b(b − 3a) l −4c(2c − 6d ) + d(3d − 2c)m 6m(2m − 3) − (2m + 4) n 2p(p − 4) + 3(5p − 2)o 7x(5 − x) + 6(x − 1) p −2y(5y − 1) − 4(2y + 3)

7 MC a What is the equivalent of 3(a + 2b) + 2(2a − b)?a 5a + 6b b 7a + 4b C 5(3a + b) d 7a + 8b

b What is the equivalent of −3(x − 2y) − (x − 5y)? a −4x + 11y b −4x − 11y C 4x + 11y d 4x + 7y

c What is the equivalent of 2m(n + 4) + m(3n − 2)? a 3m + 4n − 8 b 5mn + 4m C 5mn + 10m d 5mn + 6m

8 WE13 Expand and simplify each of the following expressions.a (a + 2)(a + 3) b (x + 4)(x + 3) c (y + 3)(y + 2)d (m + 4)(m + 5) e (b + 2)(b + 1) f (p + 1)(p + 4)g (a − 2)(a + 3) h (x − 4)(x + 5) i (m + 3)(m − 4)j (y + 5)(y − 3) k (y − 6)(y + 2) l (x − 3)(x + 1)m (x − 3)(x − 4) n (p − 2)(p − 3) o (x − 3)(x − 1)

9 Expand and simplify each of the following.a (2a + 3)(a + 2) b (3m + 1)(m + 2) c (6x + 4)(x + 1)d (c − 6)(4c − 7) e (7 − 2t)(5 − t) f (1 − x)(9 − 2x)g (2 + 3t)(5 − 2t) h (7 − 5x)(2 − 3x) i (5x − 2)(5x − 2)

10 Expand and simplify each of the following.a (x + y)(z + 1) b (p + q)(r + 3) c (2x + y)(z + 4)d (3p + q)(r + 1) e (a + 2b)(a + b) f (2c + d )(c − 3d )g (x + y)(2x − 3y) h (4p − 3q)(p + q) i (3y + z)(x + z)j (a + 2b)(b + c) k (3p − 2q)(1 − 3r) l (7c − 2d )(d − 5)m (4x − y)(3x − y) n (p − q)(2p − r) o (5 − 2j)(3k + 1)

11 MC a The equivalent of (x + 7)(x − 2) is:a x2 + 5x − 14 b 2x + 5C x2 − 5x − 14 d x2 + 5x + 14

b What is the equivalent of (4 − y)(7 + y)?a 28 − y2 b 28 − 3y + y2

C 28 − 3y − y2 d 11 − 2yc The equivalent of (2p + 1)(p − 5) is:

a 2p2 − 5 b 2p2 − 11p − 5

C 2p2 − 9p − 5 d 2p2 − 6p − 5

underSTandIng

12 Expand the following expressions using the FOIL method, then simplify.a (x + 3)(x − 3) b (x + 5)(x − 5) c (x + 7)(x − 7)d (x − 1)(x + 1) e (x − 2)(x + 2) f (2x − 1)(2x + 1)

Can you see a pattern? If so, explain. 13 Expand the following expressions using the FOIL method, then simplify.

a (x + 1)(x + 1) b (x + 2)(x + 2) c (x + 8)(x + 8)d (x − 3)(x − 3) e (x − 5)(x − 5) f (x − 9)(x − 9)Can you see a pattern? If so, explain.


SAMPLE E

VALUATIO

N ONLY

number and algebra


14 Simplify the following expressions.a 2.1x (3x + 4.7y) − 3.1y(1.4x + y)

b (2.1x − 3.2y) (2.1x + 3.2y)

c (3.4x + 5.1y)2

15 For the box shown at right, find in expanded form: a the total surface area b the volume.

reaSOnIng

16 For each of the following shapes: i write down the area in factor form ii expand and simplify the expressioniii discuss any limitations on the value of x.

a (x + 3y) m

(4x – y) m

b

(x + 5) cm

(2x – 1) cm

17 Show that(a − x)(a + x) − 2(a − x)(a − x) − 2x(a − x)= −(a − x)2.

PrOblem SOlvIng

18 Three students gave the following incorrect answers when expanding −5(3x − 20). i −5x − 20

ii −8x + 25

iii 15x − 100

a Explain the errors made by each student.

b What is the correct answer?

19 In a test, a student expanded brackets and obtained the following answers. Identify and correct the student’s errors and write the correct expansions.a −2(a − 5) = −2a −10

b 2b(3b − 1) = 6b2 − 1

c −2(c − 4) = 2c + 8

20 Shown below are three students’ attempts at expanding (3x + 4)(2x + 5).

STUDENT A STUDENT B

4x + 3

3x – 1

x

(continued)


SAMPLE E

VALUATIO

N ONLY

number and algebra

68 Maths Quest 9

STUDENT C

a Which student’s work was correct?b Copy each of the incorrect answers into your workbook and correct the mistakes in

each as though you were the teacher of these students. 21 Three incorrect expansions of (x + 8) (x − 3) are shown below. For each incorrect

expansion, explain what mistake has been made.a x2 + 11x + 24 b x2 + 5x + 24 c x2 − 3x

22 Rectangular floor mats have an area of (x2 + 2x − 15) cm2. a The length of the mat is (x + 5) cm. Find an expression for the width of the mat.b If the length of the mat is 70 cm, what is the width?c If the width of the mat is 1 m, what is the length?

3.6 Expansion patternsDifference of two squares • Look at the expansion of (x + 3)(x − 3).

(x + 3)(x − 3) = x2 − 3x + 3x − 9= x2 − 9

Compare it with

(x + 5)(x − 5) = x2 − 5x + 5x − 25= x2 − 25

In both cases the middle terms cancel each other out, leaving just two terms. There seems to be a pattern, so we can take the short cut:

(x + 10)(x − 10) = x2 − 102

= x2 − 100

The terms x2 and 100 are both perfect squares, so an expression such as x2 − 100 is called a difference of two squares. (Remember, difference means subtraction.) It is important to recognise this type of expression and remember its name.

Difference of two squares: (a + b)(a − b) = a2 − b2

Does this rule work if a = 7 and b = 3? Let’s check.(7 + 3)(7 − 3) = 10 × 4

= 4072 − 32 = 49 − 9

= 40The rule works. Try some more examples yourself.


SAMPLE E

VALUATIO

N ONLY

number and algebra


Use the difference of two squares rule to expand and simplify each of the following.a (x + 8)(x − 8) b (6 − x)(6 + x)c (2x − 3)(2x + 3) d (3x + 5)(5 − 3x)

THInK WrITe

a 1 Write the factorised expression. a (x + 8)(x − 8)

2 This expression is in the form(a + b)(a − b), so apply the rule a2 − b2.

= x2 − 82

= x2 − 64

b 1 Write the factorised expression. b (6 − x)(6 + x)

2 The difference of two squares rule can be used. Expand using the formula.

= 62 − x2

= 36 − x2

Note: 36 − x2 is not the same as x2 − 36.

c 1 Write the factorised expression. c (2x − 3)(2x + 3)

2 The difference of two squares rule can be used. Expand using the formula.

= (2x)2 − 32

= 4x2 − 9

d 1 Write the factorised expression. d (3x + 5)(5 − 3x)

2 The difference of two squares rule can be used by rearranging the terms, since 3x + 5 = 5 + 3x. Expand using the formula.

(5 + 3x)(5 − 3x) = 52 − (3x)2

= 25 − 9x2

The perfect square of a binomial • A perfect square is the square of a whole number. For example, 1, 4, 9, 16 . . . and so on

are all perfect squares since 1 = 1 × 1 = 12, 4 = 2 × 2 = 22, 9 = 3 × 3 = 32, 16 = 4 × 4 = 42 and so on.

• Similarly, (x + 3)2 is a perfect square since it is equivalent to (x + 3)(x + 3), or x + 3 multiplied by itself. It is called the ‘perfect square of a binomial.’

• Consider a picture of (x + 3)2. What shape is it?x 3

3

x x2 3x

93x

The area is given byx2 + 3x + 3x + 9 = x2 + 6x + 9

The large square is made up of two smaller squares and two rectangles, and because the rectangles are the same size (x × 3), there is a short cut to the answer.The fi rst term is x2,the third term is 32 andthe middle term is x × 3 × 2.



SAMPLE E

VALUATIO

N ONLY

number and algebra

70 Maths Quest 9

Once again there is a pattern to remember. It is called the ‘square of a binomial’.

Square of a binomial: (a + b)2 = a2 + 2ab + b2

So (x + 5)2 = x2 + 2 × x × 5 + 52

= x2 + 10x + 25 • Does this rule work if a = 7 and b = 3? Let’s check.

(7 + 3)2 = 102

= 100(7 + 3)2 = 72 + 2 × 7 × 3 + 32

= 49 + 42 + 9 = 100The rule works.

• Try to remember this rule in words:‘The square of a binomial equals the square of the fi rst term plus the square of the second term plus twice the product of the two terms.’

• Note that (a − b)2 = a2 − 2ab + b2

since a2 = a2

(−b)2 = b2

and 2 × a × −b = −2ab.

Square of a binomial: (a ± b)2 = a2 ± 2ab + b2

Use (a ± b)2 = a2 ± 2ab + b2 to expand and simplify the following.a (x + 1)(x + 1) b (x − 2)2

c (2x + 5)2 d (4x − 5y)2

THInK WrITe

a 1 This is the square of a binomial. a (x + 1)(x + 1)

2 Apply the formula for perfect squares: (a + b)2 = a2 + 2ab + b2.

= x2 + 2 × x × 1 + 12

= x2 + 2x + 1

b 1 This is the square of a binomial. b (x − 2)2

2 Apply the formula for perfect squares: (a − b)2 = a2 − 2ab + b2.

= (x − 2)(x − 2) = x2 − 2 × x × 2 + 22

= x2 − 4x + 4c 1

2

This is the square of a binomial.

Apply the formula for perfect squares:(a + b)2 = a2 + 2ab + b2.

c (2x + 5)2

= (2x)2 + 2 × 2x × 5 + 52

= 4x2 + 20x + 25

d 1

2

This is the square of a binomial.

Apply the formula for perfect squares:(a − b)2 = a2 − 2ab + b2.

d (4x − 5y)2

= (4x)2 − 2 × 4x × 5y + (5y)2

= 16x2 − 40xy + 25y2



SAMPLE E

VALUATIO

N ONLY

number and algebra


Exercise 3.6 Expansion patterns IndIvIdual PaTHWaYS

⬛ PraCTISeQuestions:1–10

⬛ COnSOlIdaTeQuestions:1–12

⬛ maSTerQuestions:1–13

FluenCY

1 WE14 Use the difference of two squares rule to expand and simplify each of the following.a (x + 2)(x − 2) b (y + 3)(y − 3) c (m + 5)(m − 5)

d (a + 7)(a − 7) e (x + 6)(x − 6) f (p − 12)(p + 12)

g (a + 10)(a − 10) h (m − 11)(m + 11)

2 WE14 Use the difference of two squares rule to expand and simplify each of the following.a (2x + 3)(2x − 3) b (3y − 1)(3y + 1) c (5d − 2)(5d + 2)

d (7c + 3)(7c − 3) e (2 + 3p)(2 − 3p) f (d − 9x)(d + 9x)

g (5 − 12a)(5 + 12a) h (3x + 10y)(3x − 10y) i (2b − 5c)(2b + 5c)

j (10 − 2x)(2x + 10)

3 WE15 Use the square of a binomial rule to expand and simplify each of the following.a (x + 2)(x + 2) b (a + 3)(a + 3) c (b + 7)(b + 7) d (c + 9)(c + 9)

e (m + 12)2 f (n + 10)2 g (x − 6)2 h (y − 5)2

i (9 − c)2 j (8 + e)2 k 2(x + y)2 l (u − v)2

4 WE15 Use the perfect squares rule to expand and simplify each of the following.a (2a + 3)2 b (3x + 1)2 c (2m − 5)2 d (4x − 3)2

e (5a − 1)2 f (7p + 4)2 g (9x + 2)2 h (4c − 6)2

i (3 + 2a)2 j (5 + 3p)2 k (2 − 5x)2 l (7 − 3a)2

m (9x − 4y)2 n (8x − 3y)2

underSTandIng

5 A square has a perimeter of 4x + 12. What is its area?

6 Francis has fenced off a square in her paddock for spring lambs. The area of the paddock is 9x2 + 6x + 1 m2.By using pattern recognition, fi nd the side length, in terms of x, of the paddock.

reaSOnIng

7 Show that a2 − b2 = (a + b)(a − b) is true for each of the following.a a = 5, b = 4 b a = 9, b = 1c a = 2, b = 7 d a = −10, b = −3

8 Anne has a square bedroom. Her sister Beth has a room that is 1 m shorter in length than Anne’s room, but 1 m wider. a Show that Anne has the larger bedroom.

b How much bigger is Anne’s bedroom compared to Beth’s bedroom?

reFleCTIOnHow could you represent (x − 3)2 on a diagram?


doc-10821doc-10821doc-10821


SAMPLE E

VALUATIO

N ONLY

number and algebra

72 Maths Quest 9

PrOblem SOlvIng

9 A large square has been subdivided into two squares and two rectangles.a Write formulas for the areas of these four pieces, using

the dimensions a and b marked on the diagram. b Write an equation that states that the area of the large

square is equal to the combined area of its four pieces. Do you recognise this equation?

10 Expand each of the following pairs of expressions.a i (x − 3)2 and (3 − x)2

ii (x − 15)2 and (15 − x)2

iii (3x − 7)2 and (7 − 3x)2

b What do you notice about the answers to the pairs of expansions above?c Explain how this is possible.

11 Expand each of the following pairs of expressions.a i (x − 4)(x + 4) and (4 − x)(4 + x) ii (x − 11)(x + 11) and (11 − x)(11 + x) iii (2x − 9)(2x + 9) and (9 − 2x)(9 + 2x) b What do you notice about the answers to the pairs of expansions above?c Explain how this is possible.

12 a Expand (10k + 5)2.b Show that (10k + 5)2 = 100k(k + 1) + 25c Using part b, evaluate 252 and 852.

13 The expansion of perfect squares (a + b)2 = a2 + 2ab + b2

and (a − b)2 = a2 − 2ab + b2

can be used to simplify some arithmetic calculations. For example: 972 = (100 − 3)2

= 1002 − 2 × 100 × 3 + 32

= 9409Use this method to calculate the following.a 1032

b 622

c 9972

d 10122

e 532

f 982

CHallenge 3.2CHallenge 3.2CHallenge 3.2CHallenge 3.2CHallenge 3.2CHallenge 3.2

b

b

a

a


SAMPLE E

VALUATIO

N ONLY

number and algebra


3.7 Further expansionsExpanding several sets of brackets • We need to take care with signs when one expression is subtracted from another.

For example, x2 + x + 2 − (x − 2) (x + 2) = x2 + x + 2 − (x2 − 4) (Remember — all of (x − 2) (x + 2) is being

subtracted.) = x2 + x + 2 − x2 + 4 = x + 6

Expand and simplify each of the following expressions.a (x + 3)(x + 4) + 4(x − 2) b (x − 2)(x + 3) − (x − 1)(x + 2)

THInK WrITe

a 1 Expand each set of brackets. a (x + 3)(x + 4) + 4(x − 2)

2 Simplify by collecting like terms. = x2 + 4x + 3x + 12 + 4x − 8 = x2 + 11x + 4

b 1 Expand and simplify each pair of brackets. Keep the second expression in a bracket.

b (x − 2)(x + 3) − (x − 1)(x + 2)

= x2 + 3x − 2x − 6 − (x2 + 2x − x − 2)

2 Subtract all of the second result from the fi rst result. Remember that −(x2 + x − 2) = −1(x2 + x − 2).

= x2 + x − 6 − (x2 + x − 2) = x2 + x − 6 − x2 − x + 2

3 Simplify by collecting like terms. = −4

Exercise 3.7 Further expansions IndIvIdual PaTHWaYS

⬛ PraCTISeQuestions:1–16, 31, 34

⬛ COnSOlIdaTeQuestions:10–22, 32–34


FluenCY

WE16 Expand and simplify each of the following expressions. 1 (x + 3)(x + 5) + (x + 2)(x + 3) 2 (x + 4)(x + 2) + (x + 3)(x + 4) 3 (x + 5)(x + 4) + (x + 3)(x + 2) 4 (x + 1)(x + 3) + (x + 2)(x + 4) 5 (p − 3)(p + 5) + (p + 1)(p − 6) 6 (a + 4)(a − 2) + (a − 3)(a − 4) 7 (p − 2)(p + 2) + (p + 4)(p − 5) 8 (x − 4)(x + 4) + (x − 1)(x + 20) 9 (y − 1)(y + 3) + (y − 2)(y + 2) 10 (d + 7)(d + 1) + (d + 3)(d − 3) 11 (x + 2)(x + 3) + (x − 4)(x − 1) 12 (y + 6)(y − 1) + (y − 2)(y − 3) 13 (x + 2)2 + (x − 5)(x − 3) 14 (y − 1)2 + (y + 2)(y − 4)


reFleCTIOnOn a diagram how would you show (m + 2)(m + 3) − (m + 2)(m + 4)?



SAMPLE E

VALUATIO

N ONLY

number and algebra

74 Maths Quest 9

15 (p + 2)(p + 7) + (p − 3)2 16 (m − 6)(m − 1) + (m + 5)2

17 (x + 3)(x + 5) − (x + 2)(x + 5) 18 (x + 5)(x + 2) − (x + 1)(x + 2) 19 (x + 3)(x + 2) − (x + 4)(x + 3) 20 (m − 2)(m + 3) − (m + 2)(m − 4) 21 (b + 4)(b − 6) − (b − 1)(b + 2) 22 (y − 2)(y − 5) − (y + 2)(y + 6) 23 (p − 1)(p + 4) − (p − 2)(p − 3) 24 (x + 7)(x + 2) − (x − 3)(x − 4) 25 (c − 2)(c − 1) − (c + 6)(c + 7) 26 (f − 7)(f + 2) − (f + 4)(f + 5) 27 (m + 3)2 − (m + 4)(m − 2) 28 (a − 6)2 − (a − 2)(a − 3) 29 (p − 3)(p + 1) − (p + 2)2 30 (x + 5)(x − 4) − (x − 1)2

reaSOnIng

31 Show that (p − 1)(p + 2) + (p − 3)(p + 1) = 2p2 − p − 5. 32 Show that (x + 2)(x − 3) − (x + 1)2 = −3x − 7.

PrOblem SOlvIng

33 a Show that (a2 + b2)(c2 + d2) = (ac − bd)2 + (ad + bc)2.b Using part a, write (22 + 12)(32 + 42) as the sum of two squares and evaluate.

34 a Expand (x2 + x − 1)2.b Show that (x2 + x − 1)2 = (x − 1)x(x + 1)(x + 2) + 1.c i Evaluate 4 × 3 × 2 × 1 + 1. ii What is the value of x if 4 × 3 × 2 × 1 + 1 = (x − 1)x(x + 1)(x + 2) + 1?

35 a Expand (a + b) (d + e).b Expand (a + b + c) (d + e + f ). Draw a diagram to illustrate your answer.

3.8 The highest common factor • Factorising a number or an expression means writing it as a product. For example, 35 is

factorised by writing it as 7 × 5, the product of 7 and 5.The factors of 35 are 7 and 5:

35 ÷ 7 = 535 ÷ 5 = 7

• 5(x + 2) is also a factorised expression. (Remember that it means 5 × (x + 2).)The factors of 5(x + 2) are 5 and (x + 2):

5(x + 2) ÷ 5 = (x + 2)5(x + 2) ÷ (x + 2) = 5

• In section 3D, factorised expressions were expanded as5(x + 2) = 5x + 10

This section examines the opposite process. We will start with expanded expressions and try to factorise them.

Complete the following factorisations.a 15x = 5x × . . . b 15x2 = 3x × . . . c −18ab2 = 6a × . . .THInK WrITe

a Divide 15x by 5x. a 15x5x

= 3

So 15x = 5x × 3.



SAMPLE E

VALUATIO

N ONLY

number and algebra


b Divide 15x2 by 3x. b 15x2

3x= 5x

So 15x2 = 3x × 5x.

c Divide −18ab2 by 6a. c −18ab2

6a= −3b2

So −18ab2 = 6a × −3b2.

• The highest common factor (HCF) of two or more numbers is the largest factor that divides into all the numbers with no remainder. For example, the factors of 12 are 1, 2, 3, 4, 6 and 12. The factors of 16 are 1, 2, 4, 8 and 16. The HCF of 12 and 16 is 4.

• The simplest way to factorise an expression is to take out the highest common factor or HCF.

Determine the highest common factor for each of the following pairs of terms.a 25a2b and 10ab b 3xy and −3xz

THInK WrITe

a 1 Write 25a2b in expanded form. a 25a2b = 5 × 5 × a × a × b

2 Write 10ab in expanded form. 10ab = 2 × 5 × a × b

3 Write the HCF. HCF = 5ab

b 1 Write 3xy in expanded form. b 3xy = 3 × x × y

2 Write −3xz in expanded form. −3xz = −1 × 3 × x × z

3 Write the HCF. HCF = 3x

Factorising expressions by fi nding the highest common factor • To factorise an expression such as 4xy + 12x, fi nd the highest common factor of both

terms. The HCF for 4xy and 12x is 4x, so 4xy + 12x = 4x × (. . .) where the brackets contain a

binomial.4xy4x

= y and 12x4x

= 3

So 4xy + 12x = 4x × (y + 3)= 4x(y + 3)

The answer can be checked by expanding the brackets.

Factorise each expression by fi rst fi nding the HCF.a 5x + 15y b −14xy − 7y c 6x2y + 9xy2

THInK WrITe

a 1 HCF = 5. Write 5 multiplied by brackets.

a 5x + 15y = 5( )

2 Divide each term by 5 to get the

binomial. 5x5

= x, 15y5

= 3y

= 5(x + 3y)




SAMPLE E

VALUATIO

N ONLY

number and algebra

76 Maths Quest 9

3 Check the answer by expanding. 5(x + 3y) = 5x + 15y (correct)

b 1 HCF = 7y or −7y, but −7y works neatly.

b −14xy − 7y

2−14xy−7y

= 2x, −7y−7y

= 1 = −7y(2x + 1)

3 Check the answer by expanding. −7y(2x + 1) = −14xy −7y (correct)

c 1 HCF = 3xy c 6x2y + 9xy2

26x2y3xy

= 2x, 9xy2

3xy= 3y = 3xy(2x + 3y)

3 Check the answer by expanding. 3xy(2x + 3y) = 6x2y + 9xy2 (correct)

Exercise 3.8 The highest common factor IndIvIdual PaTHWaYS

⬛ PraCTISeQuestions:1a–f, 2a–i, 3, 4a–i, 5a–i, 6a–i, 7a–i, 8a–i, 9

⬛ COnSOlIdaTeQuestions:1d–i, 2e–k, 3, 4g–n 5g–n, 6g–o, 7g–o, 8g–o, 9, 11

⬛ maSTerQuestions:1g–l, 2i–p, 3, 4i–r, 5i–o, 6i–r, 7j–o, 8j–o, 9–12

FluenCY

1 WE17 Complete each of the following factorisations by writing in the missing factor.a 8a = 4 × . . . b 8a = 2a × . . . c 12x2 = 4x × . . .d −12x2 = 3x2 × . . . e 3x2 = x × . . . f 15a2b = ab × . . .g 12x = −4 × . . . h 10mn = 10n × . . . i 10mn = −10 × . . .j a2b2 = ab × . . . k 30x2 = 10x × . . . l −15mn2 = −3m × . . .

2 WE18 Find the highest common factor (HCF) of each of the following.a 4 and 12 b 6 and 15 c 10 and 25 d 24 and 32

e 12, 15 and 21 f 25, 50 and 200 g 17 and 23 h 6a and 12ab

i 14xy and 21xz j 60pq and 30q k 50cde and 70fgh l 6x2 and 15x

m 6a and 9c n 5ab and 25 o 3x2y and 4x2z p 4k and 6

3 MC What is 5m the highest common factor of?a 2m and 5m b 5m and m

C 25mn and 15lm d 20m and 40m

4 WE19 Factorise each of the following expressions.a 4x + 12y b 5m + 15n c 7a + 14bd 7m − 21n e −8a − 24b f 8x − 4yg −12p − 2q h 6p + 12pq + 18q i 32x + 8y + 16zj 16m − 4n + 24p k 72x − 8y + 64pq l 15x2 − 3ym 5p2 − 20q n 5x + 5 o 56q + 8p2

p 7p − 42x2y q 16p2 + 20q + 4 r 12 + 36a2b − 24b2

reFleCTIOnHow do you � nd the factors of terms within algebraic expressions?


doc-10822doc-10822doc-10822

doc-10823doc-10823doc-10823


SAMPLE E

VALUATIO

N ONLY

number and algebra


5 Factorise each expression.a 9a + 21b b 4c + 18d2 c 12p2 + 20q2 d 35 − 14m2n e 25y2 − 15x f 16a2 + 20bg 42m2 + 12n h 63p2 + 81 − 27y i 121a2 − 55b + 110cj 10 − 22x2y3 + 14xy k 18a2bc − 27ab − 90c l 144p + 36q2 − 84pqm 63a2b2 − 49 + 56ab2 n 22 + 99p3q2 − 44p2r o 36 − 24ab2 + 18b2c

6 Factorise the following expressions.a −x + 5 b −a + 7 c −b + 9d −2m − 6 e −6p − 12 f −4a − 8g −3n2 + 15m h −7x2y2 + 21 i −7y2 − 49zj −12p2 − 18q k −63m + 56 l −12m3 − 50x3

m −9a2b + 30 n −15p − 12q o −18x2 + 4y2

p −3ab + 18m − 21 q −10 − 25p2 − 45q r −90m2 + 27n + 54p3

7 Factorise each of the following expressions.a a2 + 5a b m2 + 3m c x2 − 6xd 14q − q2 e 18m + 5m2 f 6p + 7p2

g 7n2 − 2n h a2 − ab + 5a i 7p − p2q + pqj xy + 9y − 3y2 k 5c + 3c2d − cd l 3ab + a2b + 4ab2

m 2x2y + xy + 5xy2 n 5p2q2 − 4pq + 3p2q o 6x2y2 − 5xy + x2y 8 Factorise each of the following expressions.

a 5x2 + 15x b 10y2 + 2y c 12p2 + 4p d 24m2 − 6m e 32a2 − 4a f −2m2 + 8m g −5x2 + 25x h −7y2 + 14y i −3a2 + 9aj −12p2 − 2p k −15b2 − 5b l −26y2 − 13y m 4m − 18m2 n −6t + 36t2 o −8p − 24p2

underSTandIng

9 A large billboard display is in the shape of a rectangle as shown at right. There are 3 regions (A, B, C) with dimensions in terms of x as shown.a Determine the total area of the rectangle. Give your

answer in factorised form.b Determine the area of each region in simplest form.

reaSOnIng

10 On her recent Algebra test, Marcia wrote down her answer of (a + b) (a − b)2 to the question ‘Using factorisation, simplify the following expression a2(a − b) − b2 (a − b)’. If Marcia used difference of two squares in her solution, explain the steps she took to get her answer.

PrOblem SOlvIng

11 It has been said that, for any two numbers, the product of their LCM and HCF is the same as the product of the two numbers themselves.Show whether this is true.

12 a Factorise 36x2 − 100y2 by first taking out the common factor and then using DOTS.b Factorise 36x2 − 100y2 by first using DOTS and then taking out the common factor.c Do you get the same answers to parts a and b?

(x + 3 )

(x + 1 )

x

2x

B

A

C


SAMPLE E

VALUATIO

N ONLY

number and algebra

78 Maths Quest 9

3.9 The highest common binomial factor • When factorising, always look for the highest common factor(s) fi rst.

The common binomial factor • Common factors can be expressions. • Consider the following expression: 5(x + y) + 6b(x + y).

The factors of 5(x + y) are 5 and x + y, and the factors of 6b(x + y) are 6b and x + y. So x + y is a common factor.

5(x + y)x + y

= 5 and 6b(x + y)

x + y= 6b

5(x + y) + 6b(x + y) = (x + y)(5 + 6b)

Factorise each of the following expressions.a 5(x + y) + 6b(x + y) b 2b(a − 3b) − (a − 3b)In both these expressions it can be seen that the HCF is a binomial factor.

THInK WrITe

a 1 The HCF = (x + y). a 5(x + y) + 6b(x + y)

25(x + y)

x + y= 5,

6b(x + y)x + y

= 6b = (x + y)(5 + 6b)

b 1 The HCF = (a − 3b). b 2b(a − 3b) − (a − 3b)

2 2b(a − 3b)a − 3b

= 2b, −1(a − 3b)

a − 3b= −1 = 2b(a − 3b) − 1(a − 3b)

= (a − 3b)(2b − 1)

Factorising by grouping terms • If an algebraic expression has 4 terms and no common factor in all the terms, it may be

possible to group the terms in pairs and fi nd a common factor in each pair.

Factorise each of the following expressions by grouping the terms in pairs.a 5a + 10b + ac + 2bc b x − 3y + ax − 3ay c 5p + 6q + 15pq + 2THInK WrITe

a 1 5a + 10b = 5(a + 2b)ac + 2bc = c(a + 2b)

a 5a + 10b + ac + 2bc

2 a + 2b is a common factor. = 5(a + 2b) + c(a + 2b) = (a + 2b)(5 + c)

b 1 x − 3y = 1(x − 3y), ax − 3ay = a(x − 3y)

b x − 3y + ax − 3ay

2 Take out the common factor x − 3y. = 1(x − 3y) + a(x − 3y) = (x − 3y)(1 + a)




SAMPLE E

VALUATIO

N ONLY

number and algebra


c 1 There are no simple common factors.Write the terms in a different order.

c 5p + 6q + 15pq + 2

2 5p + 15pq = 5p(1 + 3q)6q + 2 = 2 (3q + 1)

= 5p + 15pq + 6q + 2

3 1 + 3q = 3q + 1 = 5p(1 + 3q) + 2(3q + 1) = 5p(1 + 3q) + 2(1 + 3q)

4 Factorise by taking out a binomial common factor.

= (1 + 3q)(5p + 2)

• The answers can be checked by expanding the brackets.

Exercise 3.9 The highest common binomial factor IndIvIdual PaTHWaYS

⬛ PraCTISeQuestions:1–3, 5

⬛ COnSOlIdaTeQuestions:1–5


FluenCY

1 WE20 Factorise each of the following expressions.a 2(a + b) + 3c(a + b) b 4(m + n) + p(m + n)c 7x(2m + 1) − y(2m + 1) d 4a(3b + 2) − b(3b + 2)e z(x + 2y) − 3(x + 2y) f 12p(6 − q) − 5(6 − q)g 3p2(x − y) + 2q(x − y) h 4a2(b − 3) + 3b(b − 3)i p2(q + 2p) − 5(q + 2p) j 6(5m + 1) + n2(5m + 1)

2 WE21 Factorise each of the following expressions by grouping the terms in pairs.a xy + 2x + 2y + 4 b ab + 3a + 3b + 9c xy − 4y + 3x − 12 d 2xy + x + 6y + 3e 3ab + a + 12b + 4 f ab − 2a + 5b − 10g m − 2n + am − 2an h 5 + 3p + 15a + 9ap i 15mn − 5n − 6m + 2 j 10pq − q − 20p + 2k 6x − 2 − 3xy + y l 16p − 4 − 12pq + 3qm 10xy + 5x − 4y − 2 n 6ab + 9b − 4a − 6o 5ab − 10ac − 3b + 6c p 4x + 12y − xz − 3yz q 5pr + 10qr − 3p − 6q r ac − 5bc − 2a + 10b

reaSOnIng

3 Using the method of rectangles to expand, show how a(m + n) + 3(m + n) equals (a + 3)(m + n).

PrOblem SOlvIng

4 a Expand the expression (3x + 2)(2x − 5).

reFleCTIOnHow do you factorise expressions with 4 terms?



SAMPLE E

VALUATIO

N ONLY

number and algebra

80 Maths Quest 9

b Diagrams I and II below represent the expansion of (3x + 2)(2x − 5). Examine the diagrams and explain how they work to expand the expression.

I II

c For the products below, use grid paper to construct and appropriately colour representations for their expansions. Use your diagrams to write the expanded form of each.

i (2x − 3)(x − 4) ii (x + 5)(−2x + 1)d Expand each product in part c and compare with your diagrams.

5 Five squares of increasing size and five area formulas are given below.a Use factorisation to find the side length that correlates to each area formula.b Using the area given and side lengths found, match the squares below with the

appropriate algebraic expression for their area.

I II III IV V

Aa = x2 + 6x + 9 Ab = x2 + 10x + 25 Ac = x2 + 16x + 64 Ad = x2 − 6x + 9 Ae = x2 + 12x + 36

c If x = 5 cm, use the formula given to calculate the area of each square.

6 a Write out the product 5(x + 2)(x + 3) and show that it also corresponds to the diagram below.

x

x x x x x

2

3 3 3 3 3

b Explain why 5(x + 2)(x + 3) is equivalent to (5x + 10)(x + 3). Use bracket expansion and a labelled diagram to support your answer.

c Explain why 5(x + 2)(x + 3) is equivalent to (x + 2)(5x + 15). Use bracket expansion and a labelled diagram to support your answer.


SAMPLE E

VALUATIO

N ONLY

number and algebra


7 The area formulas below relate to either squares or rectangles. i 9s2 + 48s + 64 ii 25s2 − 4 iii s2 + 4s + 3 iv 4s2 − 28s − 32a Without completing any algebraic operations, examine the formulas and determine

which belong to squares and which to rectangles. Explain your answer.b Factorise each formula and classify it as a square or rectangle. Check this against

your answer to part a.

3.10 Applications • A very important skill is to be able to convert a worded or real-life problem into an

algebra sentence or expression.

A rectangular swimming pool measures 30 m by 20 m. A path around the edge of the pool is x m wide on each side.

a Determine the area of the pool.b Write an expression for the area of the pool plus the path.c Write an expression for the area of the path.d If the path is 1.5 m wide, calculate the area of the path.

THInK WrITe/draW

a 1 Construct a drawing of the pool. a

x

x

x

30 m20 m x

2 Calculate the area of the pool. Area = length × width = 20 × 30 = 600 m2

doc-10824doc-10824doc-10824



SAMPLE E

VALUATIO

N ONLY

number and algebra

82 Maths Quest 9

b 1 Write expressions for the total length and width.

b Length = 30 + x + x = 30 + 2x

Width = 20 + x + x = 20 + 2x

2 Area = length × width Area = length × width = (30 + 2x)(20 + 2x) = 600 + 60x + 40x + 4x2

= (600 + 100x + 4x2) m2

c Find an expression for the area of the path by subtracting the area of the pool from the total area.

c Area of path = total area − area of pool = 600 + 100x + 4x2 − 600 = (100x + 4x2) m2

d Substitute 1.5 for x in the expression found for the area of the path.

d When x = 1.5,Area of path = 100(1.5) + 4(1.5)2

= 159 m2

• The algebraic expression found in part c of Worked example 22 allows us to calculate the area of the path for any given width, x.


Suppose that the page of a typical textbook is 24 cm high by 16 cm wide. The page has margins at the top and bottom of x cm and on the left and right of y cm.a Write an expression for the height of the page

that is inside the margins.b Write an expression for the width of the page

that is inside the margins.c Write an expression for the area of the page that

is inside the margins.d Show that if the margins on the left and right are

doubled, then the area available to be printed is reduced by (48y − 4xy) cm2.

e If the margins at the top and the bottom of the page are 1.5 cm and the margins on the left and right of the page are 1 cm, calculate the size of the area that is available to be printed.

THInK WrITe/draW

a 1 Construct a drawing, showing the key dimensions.

a

x

y

16 cm

24 cmy

x

WOrKed eXamPle 23WOrKed eXamPle 23WOrKed eXamPle 23WOrKed eXamPle 23

Graeme LOFTS Merrin J. EVERGREEN

AUSTRALIAN CURRICULUM EDITION

999999999AUSTRALIAN CURRICULUM EDITION

FOR VICTORIA

sciencequest

scienceGraeme LOFTS Merrin J. EVERGREEN

sciencescienceGraeme LOFTS Merrin J. EVERGREEN

questquestsciencescience

Graeme LOFTS Merrin J. EVERGREEN

sciencescienceGraeme LOFTS Merrin J. EVERGREEN

questquestquestsciencequestquest

science

SAMPLE E

VALUATIO

N ONLY

number and algebra


2 The real height (24 cm) is effectively reduced by x cm at the top and x cm at the bottom.

Height = 24 − x − x= (24 − 2x) cm

b The width (16 cm) is reduced by y cm on the left and y cm on the right.

b Width = 16 − y − y= (16 − 2y) cm

c The area of the page is the product of the width and height.

c Area1 = (24 − 2x)(16 − 2y)= (384 − 48y − 32x + 4xy) cm2

d 1 If the left and right margins are doubled, they become 2y and 2y, respectively. Determine the new expression for the width of the page.

d Width = 16 − 2y − 2y= (16 − 4y) cm

2 Determine the new expression for the reduced area.

Area2 = (24 − 2x)(16 − 4y)= 24 × 16 + 24 × −4y

− 2x × 16 −2x × −2y = (384 − 96y − 32x + 8xy) cm2

3 Determine the difference in area by subtracting the reduced area from the area obtained in part c.

Difference in area = Area1 − Area2 = 384 − 48y − 32x + 4xy

− (384 − 96y − 32x + 8xy) = 384 − 48y − 32x + 4xy

− 384 + 96y + 32x − 8xy = 48y − 4xySo the amount by which the area is reduced is (48y − 4xy) cm2.

e Using the area found in c above, substitute the values for x and y, where x = 1.5 and y = 1.

e Area1 = (384 − 48y − 32x + 4xy) cm2

= 384 − 48(1) − 32(1.5) + 4(1.5)(1)= 294 cm2

Exercise 3.10 ApplicationsIndIvIdual PaTHWaYS

⬛ PraCTISeQuestions:1–4, 8, 13, 14

⬛ COnSOlIdaTeQuestions:1–3, 5, 7–10, 13, 14

⬛ maSTerQuestions:1–3, 6, 7, 9–15

FluenCY

1 Answer the following for each shape.i Determine an expression for the perimeter.ii Determine the perimeter when x = 5.iii Determine an expression for the area and simplify by expanding if necessary.iv Determine the area when x = 5.a

3x

b

x + 2

c

4x – 1

d 4x

x

reFleCTIOnHow can you use algebraic skills in real-life situations?



SAMPLE E

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N ONLY

number and algebra

84 Maths Quest 9

e 3x + 1

2x

f

x – 3

5x + 2 g 3x + 5

x + 4

6x

3x

underSTandIng

2 WE22 A rectangular swimming pool measures 50 m by 25 m. A path around the edge of the pool is x m wide on each side.a Determine the area of the pool.b Write an expression for the area of the pool plus the path.c Write an expression for the area of the path.d If the path is 2.3 m wide, calculate the area of the path.e If the area of the path is 200 m2, write an equation that could be solved to find the

width of the path.

3 WE23 The page of a book is 20 cm high by 15 cm wide. The page has margins at the top and bottom of x cm and on the left and right of y cm.a Write an expression for the height of

the page that is inside the margins.b Write an expression for the width of

the page that is inside the margins.


SAMPLE E

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number and algebra


c Write an expression for the area of the page that is inside the margins.d Show that if the margins on the left and right are doubled, then the area available to

be printed on is reduced by (40y − 4xy) cm.4 A rectangular book cover is 8 cm long and 5 cm wide.

a Calculate the area of the book cover.b i If the length of the book cover is increased by v cm, write an expression for its

new length. ii If the width of the book cover is increased by v cm, write an expression for its

new width.iii Write an expression for the new area of the book cover and expand.iv Calculate the area of the book cover if v = 2 cm.

c i If the length of the book cover is decreased by d cm, write an expression for its new length.

ii If the width of the book cover is decreased by d cm, write an expression for its new width.

iii Write an expression for the new area of the rectangle and expand.iv Calculate the area of the book cover if d = 2 cm.

d i If the length of the book cover is made x times as long, write an expression for its new length.

ii If the width of the book cover is increased by x cm, write an expression for its new width.

iii Write an expression for the new area of the book cover and expand.iv Calculate the area of the book cover if x = 5 cm.

5 A square has dimensions of 5x m.a Write an expression for its perimeter.b Write an expression for its area.c i If its length is decreased by 2 m, write an expression for its new length.

ii If its width is decreased by 3 m, write an expression for its new width.iii Write an expression for its new area and expand.iv Calculate its area when x = 6 m.

6 A rectangular sign has a length of 2x cm and a width of x cm.a Write an expression for its perimeter.b Write an expression for its area.c i If its length is increased by y cm, write an expression for its new length.

ii If its width is decreased by y cm, write an expression for its new width.iii Write an expression for its new area and expand.iv Calculate its area when x = 4 cm and y = 3 cm using your expression.

7 A square has a side length of x cm.a Write an expression for its perimeter.b Write an expression for its area.c i If its side length is increased by y cm, write an expression for its new side length.

ii Write an expression for its new perimeter and expand.iii Calculate the perimeter when x = 5 cm and y = 9 cm.iv Write an expression for its new area and expand.

d Calculate the area when x = 3.2 cm and y = 4.6 cm.


SAMPLE E

VALUATIO

N ONLY

number and algebra

86 Maths Quest 9

8 A swimming pool with length (4p + 2) m and width 3p m is surrounded by a path of width p m.Obtain the following in expanded form.a An expression for the perimeter of the poolb An expression for the area of the poolc An expression for the length of the pool and pathd An expression for the width of the pool and pathe An expression for the perimeter of the pool and pathf An expression for the area of the pool and pathg An expression for the area of the pathh The area of the path when p = 2 m

reaSOnIng

9 The figure below right is a rectangle. For the purpose of this activity we are going to call this rectangle an ‘algebra rectangle’.The length of each algebra rectangle is x cm and the width is 1 cm.These algebra rectangles are put together to form larger rectangles in one of two ways.

Long algebra rectangle (3-long) Tall algebra rectangle (3-tall)

xxx

xxx11

x

x

111

111

Let’s find the perimeter of each algebra rectangle.Perimeter of 3-long algebra rectangle; P = (x + x + x + x + x + x) + (1 + 1) = 6x + 2Perimeter of 3-tall algebra rectangle; P = (x + x) + (1 + 1 + 1 + 1 + 1 + 1) = 2x + 6a Find the perimeter of each of the algebra rectangles and put your results into the

table below.

Type of algebra rectangle Perimeter

Type of algebra rectangle Perimeter

1-long 1-tall

2-long 2-tall

3-long 6x + 2 3-tall 2x + 6

4-long 4-tall

5-long 5-tall

b Can you see a pattern? What would be the perimeter of a 20-long algebra rectangle and a 20-tall algebra rectangle?

c For what values of x will the tall rectangle have a larger perimeter than the long rectangle?

4p + 2

3p

p

x1


SAMPLE E

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number and algebra


10 The Body Mass Index (B) is used as an indicator of whether or not a person is in a

healthy weight range for their height. It can be found using the formula B = m

h2, where

m is the person’s mass in kilograms and h is the person’s height in metres. Use the formula to answer the following questions, correct to 1 decimal place.a Calculate Richard’s Body Mass Index if he weighs 85 kg and is 1.75 m tall.b A person is considered to be in a healthy weight range if his or her Body Mass

Index lies in the range of 21 to 25 inclusive. Comment on Richard’s weight for a person of his height.

c Calculate the Body Mass Index, correct to 1 decimal place, for each of the following people: i Judy, who is 1.65 m tall and has a mass of 52 kg ii Karen, who is 1.78 m tall and has a mass of 79 kgiii Manuel, who is 1.72 m tall and has a mass of 65 kg.

d The Body Mass Index (BMI) range for children changes until adulthood. These graphs show the Body Mass Index for boys and girls aged 2 to 20 years. Study the graphs carefully and decide on possible age ranges (between 2 and 20) for Judy, Karen and Manuel if their BMI was in the Satisfactory range. State the age ranges in half-year intervals.

333231302928272625242322212019181716151413121110

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20Age (years)

Bod

y M

ass

Inde

x (k

g/m

2 )

Satisfactory

Underweight

Overweig

htObe

se

Girls

Boys

Body Mass Index for boys and girls 2 to 20 years

11 Can you evaluate 9972 without a calculator and in less than 90 seconds? We would be able to evaluate this using long multiplication, but it would take a fair amount of time and effort. Mathematicians are always looking for quick and simple ways of solving problems. What if we consider the expanding formula that produces the difference of two squares?

(a + b)(a − b) = a2 − b2

Adding b2 to both sides gives (a + b)(a − b) + b2 = a2 − b2 + b2.Simplifying and swapping sides gives a2 = (a + b)(a − b) + b2.


SAMPLE E

VALUATIO

N ONLY

number and algebra

88 Maths Quest 9

We can use this new formula and the fact that multiplying by 1000 is an easy operation to evaluate 9972.a If a = 997, what should we make the value of b so that (a + b) equals 1000?b Substitute these a and b values into the formula to

evaluate 9972.c Try this method to evaluate the following.

i 9952 ii 9902

PrOblem SOlvIng

12 A new game has been created by students for the school fair. To win the game you need to hit the target with 5 darts in the shaded region.

x

R

a Write an expression for the area of the game board (the shaded region).b If R = 7.5 cm and x = 4 cm, find the area of the shaded region.

c Show that R = A + x2

π by transposing the formula found in part a.

d If A = 80 cm2 and x = 3 cm, calculate the value of R.e The students found that the best size for the game board is when R = 10 and x = 5.

What is the percentage of the total board that is shaded? 13 A picture of a mobile phone is shown at right (dimensions are in cm).

a Use the information in the picture to write an expression for the area of: i the viewing screenii the front face of the phone.

b A friend has a phone that is 4 cm longer but 1 cm narrower. Write expressions, in expanded form, for: i the length and width of your friend’s phone ii the area of your friend’s phone.

c If the phone pictured is 5 cm wide, use your answer to part b to find the area of the front face of your friend’s phone.

14 A proposed new flag for Australian schools will have the Australian flag in the top left-hand corner. The dimensions given are in metres.

x – 2

2x

x

x

x + 1

x + 7x + 2

x


SAMPLE E

VALUATIO

N ONLY

number and algebra


a Write an expression, in factorised form, for the area of:i the Australian fl agii the proposed fl ag.

b Use the answers from part a to write the area of the Australian fl ag as a fraction of the school fl ag and then simplify this fraction.

c Use the fraction from part b to express the area of the Australian fl ag as a percentageof the proposed school fl ag.

d Using the formula for the percentage of area taken up by the Australian fl ag, fi nd the percentages for the following school fl ag widths.i 4 m ii 4.5 m iii 4.8 m

e If the percentage of the school fl ag taken up by the Australian fl ag measures the importance a school places on Australia, what can be said for the three fl ags?

15 Cubic expressions are the expanded form of expressions with three linear factors. The expansion process for three linear factors is called trinomial expansion. a Explain how the area model can be altered to show that binomial expansion

can also be altered to become a model for trinomial expansion.b Investigate the powers in cubic expressions by expanding the following expressions.

i (3x + 2)(–x + 1)(x + 1) ii (x + 5)(2x − 2)(4x − 8)iii (3 − x)(x + 8)(5 − x)

c Describe the patterns in the powers in a cubic expression.doc-10825doc-10825doc-10825


SAMPLE E

VALUATIO

N ONLY


SAMPLE E

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N ONLY


ONLINE ONLY 3.11 ReviewThe Maths Quest Review is available in a customisable format for students to demonstrate their knowledge of this topic.

The Review contains:• Fluency questions — allowing students to demonstrate the

skills they have developed to effi ciently answer questions using the most appropriate methods

• Problem Solving questions — allowing students to demonstrate their ability to make smart choices, to model and investigate problems, and to communicate solutions effectively.

A summary of the key points covered and a concept map summary of this topic are available as digital documents.

Review questionsDownload the Review questions document from the links found in your eBookPLUS.

Link to assessON for questions to test your readiness FOr learning, your progress aS you learn and your levels OF achievement.

assessON provides sets of questions for every topic in your course, as well as giving instant feedback and worked solutions to help improve your mathematical skills.

www.assesson.com.au

www.jacplus.com.au

int-0699int-0699int-0699

int-0685int-0685int-0685

int-3203int-3203int-3203

LanguageIt is important to learn and be able to use correct mathematical language in order to communicate effectively. Create a summary of the topic using the key terms below. You can present your summary in writing or using a concept map, a poster or technology.

binomialcoeffi cientcommon factorconstantdifference of two squaresdistributive law

evaluateexpandexpressionfactoriseFOILhighest common factor

pronumeralsimplifysubstitutiontermstrinomialvariable

The story of mathematicsis an exclusive Jacaranda video series that explores the history of mathematics and how mathematics helped shape the world we live in today.

One small step for man. . . (eles-1690) explains the way in which algebra was used in humanity’s quest to travel to the moon, and how without mathematics the mission would have been destined to fail.

number and algebra


SAMPLE E

VALUATIO

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number and algebra

92 Maths Quest 9

<InveSTIgaTIOn> FOr rICH TaSK Or <number and algebra> FOr PuZZleInveSTIgaTIOn

rICH TaSK

Quilt squares


SAMPLE E

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N ONLY

number and algebra


number and algebra

Trace or copy the basic block at right onto a sheet of paper. Repeat this process until you have nine blocks. Colour in each section and cut out all nine blocks.

1 Place all nine blocks together to form a 3 × 3 square. In the space provided below, draw and colour a scaled diagram of your result.

2 Describe the feature created by arranging the blocks in the manner described in question 1. Observe the shapes created by the different colours.

Medini sold her design to a company that now manufactures quilts made from 100 of these blocks. Each quilt covers an area of 1.44 m2. Each row and column has the same number of blocks. Answer the following, ignoring any seam allowances.

3 Calculate the side length of each square block.

4 If the entire quilt has an area of 1.44 m2, what is the area of each block?

5 Determine the dimensions of the yellow, black and white pieces of fabric of each block.

6 Calculate the area of the yellow, black and white pieces of fabric of each block.

7 Determine the total area of each of the three different colours required to construct this quilt.

8 Due to popular demand, the company that manufactures these quilts now makes them in different sizes. All that the customer is required to do is to specify the approximate quilt area, the three colours required and the number of blocks in the quilt (they must conform to a square or rectangular shape). Derive a general formula that would allow the company to quickly determine the areas of the three coloured fabrics in each block. Give an example and a diagram on a separate sheet of paper to illustrate your formula.

Diagram of basic block used in the quilt


SAMPLE E

VALUATIO

N ONLY

number and algebra

94 Maths Quest 9

<InveSTIgaTIOn> FOr rICH TaSK Or <number and algebra> FOr PuZZlenumber and algebra

COde PuZZle

Why does a giraffe have a long neck?Expand and simplify the expressions to �nd the puzzle’s answer.

–8xx 2 + x –3x 2 + 2x + 1

–x 2 + 12x –2x 2 – 6x 2x 2 + x 2x – 9 –95x 3x 2 – 5x – 1 2x 2 – x – 16 2x 2 – x + 1 6x 2 – 13x 6x + 10x 2 + 3x + 10 x 2 + 7x – 2x 2 + 5x – 2

–3x + 3–x 2 + 19 x 2 + 2–5x 2 + x – 3 –6x + 3 –33x + 9x 2 + 8x – 18x 2 – x 5x + 3–x – 1

= 2(x + 3) + 3(x – 1)

=

= (x – 3)(x + 2) + 6

=

= x (x – 4) + 2(2x + 1)

=

= (x + 5)(x – 2) + 2(x + 4)

=

= (x – 3)(x + 3) – x (x – 2)

=

= 5(x + 1) – 2(x – 2)

=

= (x + 2)(x – 2) + (x + 3)(x – 4)

=

= (x – 4)(x + 3) – (x + 5)(x – 3)

=

= 5(x 2 + 3x + 2) – 4(x 2 + 3x)

=

= (x + 2) 2 + 3(x – 2)

=

= (x – 1)(x + 6) + 3(x – 4)

=

= 2(x – 7)(x + 5) + 5(x + 14)

=

= –3(x + 7) – (x + 5)(x – 8)

=

= (x – 5)(x + 5) – (x + 4)(x – 4)

=

= (2 + x)(3 – x) + (x + 1)(x + 4)

=

= (x + 2)(x – 2) – (x – 1)(x + 1)

=

= –3(x 2 + x – 2) + 5(x – 1)

=

= 5(x – 7) + 2(x 2 – 3x + 18)

=

= (x – 2)2 – (x + 1)2

=

= (x – 6)(x – 7) + 2(7x – 21)

=

= (2x – 3)(3x – 2) – 6

=

= 7(2 – 3x) + 5(4x – 3)

=

= 4(2x – 6) – 3(x – 8)

== –2(x 2 + 3x – 6) – 12

=

= 2(x 2 + 3x – 7) – 2(x 2 + 7x – 7)

=

= 2x(x – 3) + 3x(6 – x)

=

= 5x(2 – x) – 3(3x + 1)

=

= 3(x 2 – x + 1) – 2(x + 2)

=

’

A N

OS

TA

EFH

LN

CD E

T

S

E HF SH T

L TM E

I E


SAMPLE E

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number and algebra

Activities3.1 Overviewvideo• The story of mathematics: One small step for

man . . . (eles-1690)

3.2 using pronumeralsInteractivity• IP interactivity 3.2 (int-4480): Using pronumeralsdigital doc • SkillSHEET (doc-6122): Alternative expressions

used for the four operations

3.3 algebra in worded problemsInteractivity• IP interactivity 3.3 (int-4481): Algebra in worded problemsdigital docs • SkillSHEET (doc-6123): Algebraic expressions• SkillSHEET (doc-6124): Substitution

into algebraic expressions

3.4 Simplifying algebraic expressionsInteractivity• IP interactivity 3.4 (int-4482): Simplifying

algebraic expressionsdigital docs • SkillSHEET (doc-6125): Like terms• SkillSHEET (doc-6126): Collecting like terms• SkillSHEET (doc-6127): Multiplying algebraic terms• SkillSHEET (doc-6128): Dividing algebraic terms• SkillSHEET (doc-10817): Adding and subtracting

integers• SkillSHEET (doc-10818): Multiplying and dividing

integers• WorkSHEET 3.1 (doc-6138)

3.5 expanding bracketsInteractivity• IP interactivity 3.5 (int-4483): Expanding bracketsdigital docs • SkillSHEET (doc-10819): Expanding brackets• SkillSHEET (doc-10820): Expanding a pair

of brackets

3.6 expansion patternsInteractivity• IP interactivity 3.6 (int-4484): Expansion patternsdigital doc• SkillSHEET (doc-10821): Recognising expansion

patterns

3.7 Further expansionsInteractivities• Expanding brackets (int-2763) • IP interactivity 3.7 (int-4485): Further expansions

3.8 The highest common factorInteractivity• IP interactivity 3.8 (int-4486): The highest common factordigital docs • SkillSHEET (doc-10822): Finding the highest

common factor• SkillSHEET (doc-10823): Factorising by fi nding

the HCF

3.9 The highest common binomial factorInteractivity• IP interactivity 3.9 (int-4487): The highest

common binomial factordigital doc • WorkSHEET 3.2 (doc-10824): Factorising

3.10 applicationsInteractivity• IP interactivity 3.10 (int-4488): Applicationsdigital doc• WorkSHEET 3.3 (doc-10825): Expanding

3.11 reviewInteractivities• Word search (int-0699)• Crossword (int-0685)• Sudoku (int-3203)digital docs• Topic summary (doc-10781)• Concept map (doc-10794)

To access ebookPluS activities, log on to www.jacplus.com.au


SAMPLE E

VALUATIO

N ONLY

number and algebra

96 Maths Quest 9

Answerstopic 3 Algebra3.2 Using pronumerals 1 a 3 b 7 c −2 d −8 e 1 f −1 g 1

3 h 1

2 i −1

4 j −1

9

2 a 6 b 7 c 3 d −2

e 3 f −2 g −7 h −11 i −1

2 j −2

3 k 1

4 l 1

m −3 n −2 o 0 3 a i 3 ii 5 iii 8 iv 5x2

b i 3 ii −9 iii −6 iv −9m2

c i 4 ii 5 iii 5 iv −7x2

d i 4 ii 9 iii 4 iv −9b2

e i 5 ii 11 iii −4 iv −7q f i 5 ii −9 iii 5 iv −9p g i 4 ii 4 iii −2 iv −3ac h i 5 ii 5 iii 9 iv −3u i i 5 ii −1 iii 8 iv −m j i 5 ii 7 iii 14 iv −3cd2

4 a 7, 2, −8 b 8, 2, −4 c 3, 0, 0 d 0, −2, 0 e 1, 0, 2 f 2, cannot be done, −1 5 a B b D c A d B 6 a 4 b −6 c 6 d 7 e −12 f −3 7 a 7 b −40 c 7 8 a i 32 ii 250 b i 4 ii 3 9 a 15 b 10 c 18 d 45 e 30 f 10.72 g 21 h 22.4 i 50 j 171 k 4 l −30 m 32.75 n 38.33 o 34.56 10 a 12 cm2 b 12 cm 11 a 15 b Answers will vary. 12 a 5 b Answers will vary. 13 a 63 cm3 b 4 cm 14 a 10 b 10 15 a 19.44 b Answers will vary. 16 a 461.81 b 9.55 17 a 1671.33 b 2.64 18 a

d 2d + 2g g

2g d + g 2d

2d + g 0 d + 2g Magic number = 3d + 3g

b, c Answers will vary. 19 a mn b (m − 2)(n − 2) c (2m + 2n) units 20 y − (z − x)3.3 Algebra in worded problems 1 a 11 b x + 5 c y + 5 2 a 3.5 km b (x + 2) km 3 a 4 b 2.5 + t c 2.5 − y 4 a 10n, where n = number of shirts b N + 30, where N = the number of Nick’s dollars c i g − 4 ii 7g − 4 5 a i 94 ii 6x + y b i 66 ii 6p + q c i 28 ii 6x + y −6p − q 6 xy dollars

7 a xn b

x2n

8 a (30 − d ) cm b 30 − d

4 cm c

3(30 − d)

4 cm

9 a x4 b

x6

c x4

− x6

= x12

10 a 24 − c

7 hours b (c + 24 − c

7) = 6c + 24

7 11 a 1 b 8.30 pm and 9.00 pm c 5 d 24 e 7.30 pm and 8.00 pm 12 a b = 2h − 1 b 19 blocks 13 4 dots = 4 triangles 5 dots = 10 triangles 6 dots = 20 triangles 7 dots = 35 triangles

t = d!6(d − 3)!

where d! is the product of all positive integers less

than or equal to d.14 15 red vases and 10 clear vases15 Answers will vary.

16 a x = T − 3T − 1

b, c Undefined; answers will vary.

Challenge 3.1The father is 32 years old.

3.4 Simplifying algebraic expressions 1 a 9ab, −ab b 4x c −3za, −az d 2x2, −x2 e −2x2y f −x2y5, −3x2y5

g p3x2w5, −5x2p3w5 h 4y5z4x2, −2x2z4y5

2 a 7x b 11y c 19m

d 11q e 8r f 3x g 8a h 8y i 13x j 16p k 9q2 l 3x2

m 8x2 −3y n 2m2 + 2n

o 7x2 + x p 9h2 + h + 9 q −2g2 + g −12 r −5m2 + m + 15 s 12a2 + b + 4b2 t 3m + 7n2

u 12xy + 2y2 v 2ab + 5a2b w 16x2y −3xy x m2n + 11n y −7x2 + 2x2y z −3a2b −9a2 + 5b + 4 3 a B b D c A d D 4 a 6mn b 20xy c 8pq d −10xy e −12xy f 15mn g 10a2 h 20y2

i 5p2 j 7m2 k 6mnp l −6ab2

m 10m2n n −18a2b o 30x3y2 p −12p2q3

q −56c3d r 30a4b7

5 a 3x b 3m c 2y d 4m e 4m f 2x g −7x h −4m

i m2

j x3 k

4m9

l 4x3

m 1

5m n

12a

o 2yz p 35ab

2

q xy

4 r

−7yz

11


SAMPLE E

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number and algebra


6 a 40x2y2 b 56ax2y2 c 12x3y2

d x2

2y e

−5a4b

f 2

g 40a3b2 h −8a3b3 i 16a4

7 a i (0.01mp + nq) dollars ii (mp + 100nq) cents b 20 − (0.01mp + nq) dollars 8 Shirt = $12 each Trousers = $30 each 9 Explanations will vary. The correct answer is 7a2c2.

10 a i

ii

b i

ii

11 a P = x3

+ x + 5x7

b Answers will vary. c $18

12 a P = 2w + 32w7

b Answers will vary. c $72.22 13 a P = 2x + 2(4x − 7) b P = 10x − 14 c 16 d If x = 1 the rectangle will have a negative value perimeter,

which is not possible.14 a l = 4(x + 20) + 4(x + 40) + 4(x + 20) b l = 12x + 320 c $19.2015 a P1 = 10x + 12 cm P2 = 16x cm b x = 2

3.5 Expanding brackets 1 a 64 b 1100 c −20 d 693 2 a 3x + 6 b 4x + 12

3 3x 6

x 2

4 4x 12

x 3 c 4x + 4 d 7x − 7 e −3p + 6 f −x + 1 g 6b − 12 h 24m − 16 3 a x2 + 2x

x x2 2x

x 2 b a2 + 5a

a a2 5a

a 5

c 4x + x2 d 7m − m2 e 2xy + 4x f −3xy −12y g −3b + ab h −30a + 18a2

4 a 2p − 2 b 5x − 17 c −7p − 17 d −12p + 3 e 6x2 − 20x f 2m2 + 7m g 3px + 6x − 5 h 4y2 − 4y + 7 i −4p2 + 13p j 4x − 13y k 2m2 − 8m −4 l −3p2 + 10pq −1 m −30a + 10ab n 7cd − 12c2 − 5c o −2p − 17 p 3 − 3m

5 a 8x + y b 10p + 8q c 18a + 29b d 19c + 22d e 2m − 11n f 6x − 11y g 9x + 5y h −14p − 8q i −4a + 3b j 4x − y k 5p − 10q l −8c + 9d m 13x − 19y n −7p + 11q o −5a + 3b p 8c + d 6 a 2ab + 2a − 3b b 2xy + 4x − 2y c 2cd + 3c d 2pq − 2p e 5cd − 11c f 5ab − 21a − 3b g 5m h 6cd − 36c i 6m2 + 6m − 10 j 9cd − 8c k −15a2 + 2b2 − 9ab l −8c2 + 3d2 + 22cd m 12m2 − 20m − 4 n 2p2 + 7p − 6 o −7x2 + 41x − 6 p −10y2 − 6y − 12 7 a B b A c D 8 a a2 + 5a + 6 b x2 + 7x + 12 c y2 + 5y + 6 d m2 + 9m + 20 e b2 + 3b + 2 f p2 + 5p + 4 g a2 + a − 6 h x2 + x − 20 i m2 − m − 12 j y2 + 2y − 15 k y2 − 4y − 12 l x2 − 2x − 3 m x2 − 7x + 12 n p2 − 5p + 6 o x2 − 4x + 3 9 a 2a2 + 7a + 6 b 3m2 + 7m + 2 c 6x2 + 10x + 4

d 4c2 − 31c + 42 e 2t2 − 17t + 35 f 2x2 − 11x + 9 g −6t2 + 11t + 10 h 15x2 − 31x + 14 i 25x2 − 20x + 4

10 a xz + x + yz + y b pr + 3p + qr + 3q c 2xz + 8x + yz + 4y d 3pr + 3p + qr + q e a2 + 3ab + 2b2 f 2c2 − 5cd − 3d2

g 2x2 − xy − 3y2 h 4p2 + pq − 3q2

i 3xy + 3yz + xz + z2 j ab + ac + 2b2 + 2bc k 3p − 9pr − 2q + 6qr l 7cd − 35c − 2d2 + 10d m 12x2 − 7xy + y2 n 2p2 − pr − 2pq + qr o 15k + 5 − 6jk − 2j 11 a A b C c C 12 a x2 − 9 b x2 − 25 c x2 − 49 d x2 − 1 e x2 − 4 f 4x2 − 1 13 a x2 + 2x + 1 b x2 + 4x + 4 c x2 + 16x + 64 d x2 − 6x + 9 e x2 − 10x + 25 f x2 − 18x + 81 14 a 6.3x2 + 5.53xy − 3.1y2 b 4.41x2 − 10.24y2

c 11.56x2 + 34.68xy + 26.01y2

15 a Surface area = 38x2 + 14x − 6 b Volume = 12x3 + 5x2 − 3x 16 a i (x + 3y)(4x − y) ii 4x2 + 11xy − 3y2 iii x > −3y, x >

y

4

b i (2x − 1)(x + 5)

2 ii

2x2 + 9x − 52

iii x > 12

17 Answers may vary. 18 a i Student did not multiply both terms. ii Student used addition instead of multiplication. iii Student did not change negative and positive signs. b 100 − 15x 19 a −2(a − 5) = 2a − 10 is incorrect because the student did not

change the multiplied negative signs for −2 × −5 to a positive sign. The correct answer is −2a + 10.

b 2b(3b − 1) = 6b2 − 1 is incorrect because the student did not multiply −1 and 2b together. The correct answer is 6b2 − 2b.

2a + 3ba + b

3a + b2a – b

3a + 4b

–a + 2b

5p – 2q

8p + 3q

3p + 5q

p + 2q 3p – 5q2p + 3q

18m3n2

3m2n

m2 2m3n

6mn

48d2e2

8de 6de

3d2e4d


SAMPLE E

VALUATIO

N ONLY

number and algebra

98 Maths Quest 9

c −2(c − 4) = 2c + 8 is incorrect because the student has left out the negative sign when multiplying −2 and c. The correct answer is −2c + 8.

20 a Student C b Corrections to the students’ answers are shown in bold.

Student A: (3x + 4)(2x + 5) = 3x × 2x + 3x × 5 + 4 × 2x + 4 × 5 = 6x2 + 23x + 20 (Also, 29x + 20 does not equal 49x.) Student B: (3x + 4)(2x + 5) = 3x × 2x + 3x × 5 + 4 × 2x + 4 × 5 = 6x2 + 15x + 8x + 20 = 6x2 + 23x + 20 21 a Negative sign ignored b Negative sign ignored c Distributive law not used

22 a w = x2 + 2x − 15x + 5

b w = 62 cm c l = 108 cm

3.6 Expansion patterns 1 a x2 − 4 b y2 − 9 c m2 − 25 d a2 − 49 e x2 − 36 f p2 − 144 g a2 −100 h m2 −121 2 a 4x2 − 9 b 9y2 − 1 c 25d2 − 4 d 49c2 − 9 e 4 − 9p2 f d2 − 81x2

g 25 − 144a2 h 9x2 − 100y2

i 4b2 − 25c2 j 100 − 4x2

3 a x2 + 4x + 4 b a2 + 6a + 9 c b2 + 14b + 49 d c2 + 18c + 81 e m2 + 24m + 144 f n2 + 20n + 100 g x2 −12x + 36 h y2 −10y + 25 i 81 −18c + c2 j 64 + 16e + e2

k 2x2 + 4xy + 2y2 l u2 − 2uv + v2

4 a 4a2 + 12a + 9 b 9x2 + 6x + 1 c 4m2 − 20m + 25 d 16x2 − 24x + 9 e 25a2 −10a + 1 f 49p2 + 56p + 16 g 81x2 + 36x + 4 h 16c2 − 48c + 36 i 9 + 12a + 4a2 j 25 + 30p + 9p2

k 4 −20x + 25x2 l 49 − 42a + 9a2

m 81x2 − 72xy + 16y2 n 64x2 − 48xy + 9y2

5 (x2 + 6x + 9) units2

6 (3x + 1) m 7 Answers will vary. 8 Anne’s bedroom is larger by 1 m2. 9 a a2 + ab + ab + b2

b A = (a + b) × (a + b) = (a + b)2

= a2 + ab + ab + b2

= a2 + 2ab + b2

Perfect squares 10 a i x2 − 6x + 9 and 9 − 6x + x2

ii x2 − 30x + 225 and 225 − 30x + x2 iii 9x2 − 42x + 49 and 49 − 42x + 9x2

b The answers to the pairs of expansions are the same. c This is possible because when a negative number is multiplied

by itself, it becomes positive; and when expanding a perfect square, where the two expressions are the same, the negative signs cancel out and result in the same answer.

11 a i x2 − 16 and 16 − x2

ii x2 − 121 and 121 − x2 iii 4x2 − 81 and 81 − 4x2 b The answers to the pairs of expansions are the same, except

that the negative and positive signs are reversed. c This is possible because when a negative number is multiplied

by a positive number, it becomes negative; and when

expanding a DOTS, where the two expressions have different signs, the signs will be reversed.

12 a (10k + 5)2 = 100k2 + 100k + 25 b (10k + 5)2 = 100 × k × k + 100 × k + 25 = 100k(k + 1) + 25 c 252 = (10 × 2 + 5)2

Let k = 2. 252 = 100k(k + 1) + 25 = 100 × 2 × (2 + 1) + 25 = 625 852 = (10 × 8 + 5)2 Let k = 8. 852 = 100k(k + 1) + 25 = 100 × 8 × (8 + 1) + 25 = 7225 13 a 10 609 b 3844 c 994 009 d 1 024 144 e 2809 f 9604Challenge 3.2

−3(x − 1)2 4x2 − x + 4 −x2 + 4x − 1

2x2 + x + 2 3x −2x2 + 5x − 2

(x + 1)2 −4x2 + 7x − 4 3(x2 + 1)

3.7 Further expansions 1 2x2 + 13x + 21 2 2x2 + 13x + 20 3 2x2 + 14x + 26 4 2x2 + 10x + 11 5 2p2 − 3p − 21 6 2a2 − 5a + 4 7 2p2 − p − 24 8 2x2 + 19x − 36 9 2y2 + 2y − 7 10 2d2 + 8d − 2 11 2x2 + 10 12 2y2

13 2x2 − 4x + 19 14 2y2 − 4y − 7 15 2p2 + 3p + 23 16 2m2 + 3m + 31 17 x + 5 18 4x + 8 19 −2x − 6 20 3m + 2 21 −3b − 22 22 −15y − 2 23 8p − 10 24 16x + 2 25 −16c − 40 26 −14f − 34 27 4m + 17 28 −7a + 30 29 −6p − 7 30 3x − 21 31 (p − 1)(p + 2) + (p − 3)(p + 1) = (p2 + p − 2) + (p2 − 2p − 3) = p2 + p2 + p − 2p − 2 − 3 = 2p2 − p − 5 32 (x + 2)(x − 3) − (x + 1)2 = (x2 − x − 6) − (x2 + 2x + 1) = −3x − 7 33 a (a2 + b2)(c2 + d2) = a2c2 + a2d2 + b2c2 + b2d2

= (a2c2 + b2d2) + (a2d2 + b2c2) = ((ac)2 − 2abcd + (bd)2) + ((ad)2 + 2abcd + (bc)2) = (ac − bd)2 + (ad + bc)2

b a = 2, b = 1, c = 3, d = 4 (22 + 12)(32 + 42) = (2 × 3 − 1 × 4)2 + (2 × 4 −1 × 3)2

= 4 + 121 = 125 34 a (x2 + x − 1)2 = (x2 + x − 1)(x2 + x − 1) = x4 + x3 − x2 + x3 + x2 − x − x2 − x + 1 = x4 + 2x3− x2 − 2x + 1 b x4 + 2x3− x2 − 2x + 1 = x(x3 + 2x2 − x − 2) + 1 = x(x + 1)(x + 2)(x − 1) + 1 c i 4 × 3 × 2 × 1 + 1 = 25 ii x = 2


SAMPLE E

VALUATIO

N ONLY

number and algebra


35 a (a + b)(d + e) = ad + ae + bd + be b (a + b + c)(d + e + f ) = ad + ae + bd + be + cd + ce + af +

bf + cf

a b c

d ad bd cd

e ae be ce

f af bf cf

3.8 The highest common factor 1 a 2a b 4 c 3x d −4 e 3x f 15a g −3x h m i −mn j ab k 3x l 5n2

2 a 4 b 3 c 5 d 8 e 3 f 25 g 1 h 6a i 7x j 30q k 10 l 3x m 3 n 5 o x2 p 2 3 C 4 a 4(x + 3y) b 5(m + 3n) c 7(a + 2b) d 7(m − 3n) e −8(a + 3b) f 4(2x − y) g −2(6p + q) h 6(p + 2pq + 3q) i 8(4x + y + 2z) j 4(4m − n + 6p) k 8(9x − y + 8pq) l 3(5x2 − y) m 5(p2 − 4q) n 5(x + 1) o 8(7q + p2) p 7(p − 6x2y) q 4(4p2 + 5q + 1) r 12(1 + 3a2b − 2b2) 5 a 3(3a + 7b) b 2(2c + 9d2) c 4(3p2 + 5q2) d 7(5 − 2m2n) e 5(5y2 − 3x) f 4(4a2 + 5b) g 6(7m2 + 2n) h 9(7p2 + 9 − 3y) i 11(11a2 − 5b + 10c) j 2(5 − 11x2y3 + 7xy) k 9(2a2bc − 3ab − 10c) l 12(12p + 3q2 − 7pq) m 7(9a2b2 − 7 + 8ab2) n 11(2 + 9p3q2 − 4p2r) o 6(6 − 4ab2 + 3b2c) 6 a −(x − 5) b −(a − 7) c −(b − 9) d −2(m + 3) e −6(p + 2) f −4(a + 2) g −3(n2 − 5m) h −7(x2y2 − 3) i −7(y2 + 7z) j −6(2p2 + 3q) k −7(9m − 8) l −2(6m3 + 25x3) m −3(3a2b − 10) n −3(5p + 4q) o −2(9x2 − 2y2) p −3(ab − 6m + 7) q −5(2 + 5p2 + 9q) r −9(10m2 − 3n − 6p3) 7 a a(a + 5) b m(m + 3) c x(x − 6) d q(14 − q) e m(18 + 5m) f p(6 + 7p) g n(7n − 2) h a(a − b + 5) i p(7 − pq + q) j y(x + 9 − 3y) k c(5 + 3cd − d ) l ab(3 + a + 4b) m xy(2x + 1 + 5y) n pq(5pq − 4 + 3p) o xy(6xy − 5 + x) 8 a 5x(x + 3) b 2y(5y + 1) c 4p(3p + 1) d 6m(4m − 1) e 4a(8a − 1) f −2m(m − 4) g −5x(x − 5) h −7y(y − 2) i −3a(a − 3) j −2p(6p + 1) k −5b(3b + 1) l −13y(2y + 1) m 2m(2 − 9m) n −6t(1 − 6t) o −8p(1 + 3p) 9 a 2(x + 3)(4x + 1) b A = x(x + 3) C = 2(x + 3)x B = 5x2 + 17x + 6

10 Answers will vary. 11 True 12 a 4(3x − 5y)(3x + 5y) b 4(3x − 5y)(3x + 5y) c Yes, the answers are the same.

3.9 The highest common binomial factor 1 a (a + b)(2 + 3c) b (m + n)(4 + p) c (2m + 1)(7x − y) d (3b + 2)(4a − b) e (x + 2y)(z − 3) f (6 − q)(12p − 5) g (x − y)(3p2 + 2q) h (b − 3)(4a2 + 3b) i (q + 2p)(p2 − 5) j (5m + 1)(6 + n2) 2 a (y + 2)(x + 2) b (b + 3)(a + 3) c (x − 4)(y + 3) d (2y + 1)(x + 3) e (3b + 1)(a + 4) f (b − 2)(a + 5) g (m − 2n)(1 + a) h (5 + 3p)(1 + 3a) i (3m − 1)(5n − 2) j (10p − 1)(q − 2) k (3x − 1)(2 − y) l (4p − 1)(4 − 3q) m (2y + 1)(5x − 2) n (2a + 3)(3b − 2) o (b − 2c)(5a − 3) p (x + 3y)(4 − z) q (p + 2q)(5r − 3) r (a − 5b)(c − 2) 3 Answers will vary. 4 a 6x2 − 11x − 10 b Answers will vary. c Answers will vary. d i 2x2 − 11x + 12 ii –2x2 − 9x + 5 5 a Side lengtha = x + 3

Side lengthb = x + 5 Side lengthc = x + 8 Side lengthd = x − 3 Side lengthe = x + 6

b Aa = II = (x + 3)2

Ab = III = (x + 5)2

Ac = V = (x + 8)2

Ad = I = (x − 3)2

Ae = IV = (x + 6)2

c Aa = II = 64 cm2

Ab = III = 100 cm2

Ac = V = 169 cm2

Ad = I = 4 cm2

Ae = IV = 121 cm2

6 a 5(x + 2)(x + 3) = 5(x2 + 5x + 6) = 5x2 + 15x + 30

x 3 x 3 x 3 x 3 x 3

x x2 3x x2 3x x2 3x x2 3x x2 3x

2 2x 6 2x 6 2x 6 2x 6 2x 6

b 5(x + 2)(x + 3) = (5 × x + 2 × 5) (x + 3) = (5x + 10) (x + 3)

x 2 x 2 x 2 x 2 x 2

x x2 2x x2 2x x2 2x x2 2x x2 2x

3 3x 6 3x 6 3x 6 3x 6 3x 6

c 5(x + 2)(x + 3) = (5 × x + 3 × 5) (x + 2) = (5x + 15) (x + 2)

5x 15

x 5x2 15x

2 10x 30

7 a i Square, because it is a perfect square. ii Rectangle, because it is a DOTS. iii Rectangle, because it is a trinomial. iv Rectangle, because it is a trinomial. b i (3s + 8)2 ii (5s + 2)(5s − 2) iii (s + 1)(s + 3) iv 4(s2 − 7s − 8)


SAMPLE E

VALUATIO

N ONLY

number and algebra

100 Maths Quest 9

3.10 Applications 1 a i 12x ii 60 iii 9x2 iv 225 b i 4x + 8 ii 28 iii x2 + 4x + 4 iv 49 c i 16x − 4 ii 76 iii 16x2 − 8x + 1 iv 361 d i 10x ii 50 iii 4x2 iv 100 e i 10x + 2 ii 52 iii 6x2 + 2x iv 160 f i 12x − 2 ii 58 iii 5x2 −13x − 6 iv 54 g i 20x + 18 ii 118 iii 21x2 + 42x iv 735 2 a 1250 m2 b (1250 + 150x + 4x2) m2

c (150x + 4x2) m2 d 366.16 m2

e 150x + 4x2 = 200 3 a (20 − 2x) cm b (15 − 2y) cm c (300 − 40y − 30x + 4xy) cm2 d Check with your teacher. 4 a 40 cm2

b i (8 + v) cm ii (5 + v) cm iii (v2 + 13v + 40) cm2 iv 70 cm2

c i (8 − d ) cm ii (5 − d ) cm iii (40 − 13d + d2) cm2 iv 18 cm2

d i 8x cm ii (5 + x) cm iii (8x2 + 40x) cm2 iv 400 cm2

5 a 20x m b 25x2 m2

c i (5x − 2) m ii (5x − 3) m iii (25x2 − 25x + 6) m2 iv 756 m2

6 a 6x cm b 2x2 cm2

c i (2x + y) cm ii (x − y) cm iii (2x2 − xy − y2) cm2 iv 11 cm2

7 a 4x cm b x2 cm2

c i (x + y) cm ii (4x + 4y) cm iii 56 cm iv (x2 + 2xy + y2) cm2

d 60.84 cm2

8 a (14p + 4) m b (12p2 + 6p) m2

c (6p + 2) m d 5p m e (22p + 4) m f (30p2 + 10p) m2

g (18p2 + 4p) m2 h 80 m2

9 a Type of ‘algebra’ rectangle Perimeter

Type of ‘algebra’ rectangle Perimeter

1-long 2x + 2 1-tall 2x + 2

2-long 4x + 2 2-tall 2x + 4

3-long 6x + 2 3-tall 2x + 6

4-long 8x + 2 4-tall 2x + 8

5-long 10x + 2 5-tall 2x + 10

b Perimeter: long 40x + 2, tall 2x + 40 c 0 < x < 1 10 a 27.8 b Richard is not within the healthy weight range.

c i 19.1 ii 24.9 iii 22.0 d Judy: 9 to 20 years of age; Karen: 17.5 to 20 years of age;

Manuel: 13.5 to 20 years of age 11 a 3 b 994 009 c i 990 025 ii 980 100 12 a A = πR2 – x2

b A = 160.71 cm2

c Answers will vary. d R = 5.32 cm e 92.0% 13 a i Ascreen = x(x + 2) = x2 + 2x ii Aphone 1 = (x + 1)(x + 7) = x2 + 8x + 7 b i l = x + 11, w = x ii Aphone 2 = x(x + 11) = x2 + 11x c x = 4 cm, Aphone 2 = 60 cm2

14 a i x(x − 2) m2 ii x(2x) m2

b x − 2

2x

c 50(x − 2)

x%

d i 25% ii 27.8% iii 29.17% e The third flag places the most importance on Australia. 15 a Answers will vary. b i –3x3 − 2x2 + 3x + 2 ii 8x3 + 16x2 − 104x + 80 iii x3 − 49x + 120 c In a cubic expression there are four terms with descending

powers of x and ascending values of its pronumeral. For example, (x + a)3 = x3 + 3ax2 + 3a2x + a3, where the powers of x are descending and the values of a are ascending.

Investigation — Rich task 1 Solution not required 2 Answers will vary. 3 0.12 m 4 0.0144 m2

5 Yellow: 0.03 m × 0.03 m Black : 0.03 m × 0.06m White : 0.06 m × 0.06 m 6 Yellow : 0.0009 m2

Black : 0.0018 m2

White : 0.0036 m2

7 Yellow : 0.0036 m2

Black : 0.0072 m2

White : 0.0036 m2

8 Answers will vary.

Code puzzleHe can’t stand the smell of his feet.

c03Algebra.indd 100 18/05/16 2:50 PM

SAMPLE E

VALUATIO

N ONLY

c03Algebra.indd 101 18/05/16 2:50 PM

SAMPLE E

VALUATIO

N ONLY

Algebra - Homepage | Wiley · number and algebra 3.2 Using pronumerals ... a 3x2 b x + 5 + 2 C x2...

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Transcript of Algebra - Homepage | Wiley · number and algebra 3.2 Using pronumerals ... a 3x2 b x + 5 + 2 C x2...